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Chemistry library
Unit 14: lesson 1.
- Introduction to buffers
- Properties of buffers
- Henderson–Hasselbalch equation
- Common ion effect and buffers
Buffer solution pH calculations
- Methods for preparing buffers
- pH and pKa relationship for buffers
- Buffer capacity
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6.8: Buffer Solutions
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- Page ID 127254
- David Harvey
- DePauw University
Adding as little as 0.1 mL of concentrated HCl to a liter of H 2 O shifts the pH from 7.0 to 3.0. Adding the same amount of HCl to a liter of a solution that 0.1 M in acetic acid and 0.1 M in sodium acetate, however, results in a negligible change in pH. Why do these two solutions respond so differently to the addition of HCl?
A mixture of acetic acid and sodium acetate is one example of an acid–base buffer . To understand how this buffer works to limit the change in pH, we need to consider its acid dissociation reaction
\[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\]
and its corresponding acid dissociation constant
\[K_{a}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=1.75 \times 10^{-5} \label{6.1}\]
Taking the negative log of the terms in Equation \ref{6.1} and solving for pH leaves us with the result shown here.
\[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \nonumber\]
\[\mathrm{pH}=4.76+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \label{6.2}\]
You may recall that we developed these same equations in Chapter 6.6 when we introduced ladder diagrams.
Buffering occurs because of the logarithmic relationship between pH and the concentration ratio of acetate and acetic acid. Here is an example to illustrate this point. If the concentrations of acetic acid and acetate are equal, the buffer’s pH is 4.76. If we convert 10% of the acetate to acetic acid, by adding a strong acid, the ratio [CH 3 COO – ]/[CH 3 COOH] changes from 1.00 to 0.818, and the pH decreases from 4.76 to 4.67—a decrease of only 0.09 pH units.
The ratio [CH 3 COO – ]/[CH 3 COOH] becomes 0.9/1.1 = 0.818 and the pH becomes
\[\mathrm{pH}=4.76+\log (0.818)=4.67 \nonumber\]
Systematic Solution to Buffer Problems
Equation \ref{6.2} is written in terms of the equilibrium concentrations of CH 3 COOH and of CH 3 COO – . A more useful relationship relates a buffer’s pH to the initial concentrations of the weak acid and the weak base. We can derive a general buffer equation by considering the following reactions for a weak acid, HA, and the soluble salt of its conjugate weak base, NaA.
\[\begin{array}{c}{\mathrm{NaA}(s) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{A}^{-}(a q)} \\ {\mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q)} \\ {2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)}\end{array} \nonumber\]
Because the concentrations of Na + , A – , HA, H 3 O + , and OH – are unknown, we need five equations to define the solution’s composition. Two of these equations are the equilibrium constant expressions for HA and H 2 O.
\[K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \label{6.3}\]
\[K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \nonumber\]
The remaining three equations are mass balance equations for HA and Na +
\[C_{\mathrm{HA}}+C_{\mathrm{NaA}}=[\mathrm{HA}]+\left[\mathrm{A}^{-}\right] \label{6.4}\]
\[C_{\mathrm{NaA}}=\left[\mathrm{Na}^{+}\right] \label{6.5}\]
and a charge balance equation
\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\left[\mathrm{Na}^{+}\right]=\left[\mathrm{OH}^{-}\right]+\left[\mathrm{A}^{-}\right] \label{6.6}\]
Substituting Equation \ref{6.5} into Equation \ref{6.6} and solving for [A – ] gives
\[\left[\mathrm{A}^{-}\right]=C_{\mathrm{NaA} }-\left[\mathrm{OH}^{-}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \label{6.7}\]
Next, we substitute Equation \ref{6.7} into Equation \ref{6.4}, which gives the concentration of HA as
\[[\mathrm{HA}]=C_{\mathrm{HA}}+\left[\mathrm{OH}^{-}\right]-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \label{6.8}\]
Finally, we substitute Equation \ref{6.7} and Equation \ref{6.8} into Equation \ref{6.3} and solve for pH to arrive at a general equation for a buffer’s pH.
\[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{C_{\mathrm{NaA} }-\left[\mathrm{OH}^{-}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{C_{\mathrm{HA}}+\left[\mathrm{OH}^{-}\right]-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\]
If the initial concentrations of the weak acid, C HA , and the weak base, C NaA . are significantly greater than [H 3 O + ] and [OH – ], then we can simplify the general equation to the Henderson–Hasselbalch equation .
\[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{C_{\mathrm{NaA}}}{C_{\mathrm{HA}}} \label{6.9}\]
As outlined below, the Henderson–Hasselbalch equation provides a simple way to calculate the pH of a buffer, and to determine the change in pH upon adding a strong acid or strong base.
Lawrence Henderson (1878‐1942) first developed a relationship between [H 3 O + ], [HA], and [A – ] while studying the buffering of blood. Kurt Hasselbalch (1874‐1962) modified Henderson’s equation by transforming it to the logarithmic form shown in Equation \ref{6.9}. The assumptions that lead to Equation \ref{6.9} result in a minimal error in pH (<±5%) for larger concentrations of HA and A – , for concentrations of HA and A – that are similar in magnitude, and for weak acid’s with p K a values closer to 7. For most problems in this textbook, Equation \ref{6.9} provides acceptable results. Be sure, however, to test your assumptions. For a discussion of the Henderson–Hasselbalch equation, including the error inherent in Equation \ref{6.9}, see Po, H. N.; Senozan, N. M. “The Henderson–Hasselbalch Equation: Its History and Limitations,” J. Chem. Educ. 2001 , 78 , 1499–1503.
Example 6.8.1
Calculate the pH of a buffer that is 0.020 M in NH 3 and 0.030 M in NH 4 Cl. What is the pH after we add 1.0 mL of 0.10 M NaOH to 0.10 L of this buffer?
The acid dissociation constant for \(\text{NH}_4^+\) is \(5.70 \times 10^{-10}\), which is a p K a of 9.24. Substituting the initial concentrations of NH 3 and NH 4 Cl into Equation \ref{6.9} and solving, we find that the buffer’s pH is
\[\mathrm{pH}=9.24+\log \frac{0.020}{0.030}=9.06 \nonumber\]
With a pH of 9.06, the concentration of H 3 O + is \(8.71 \times 10^{-10}\) and the concentration of OH – is \(1.15 \times 10^{-5}\). Because both of these concentrations are much smaller than either \(C_{\text{NH}_3}\) or \(C_{\text{NH}_4\text{Cl}}\), the approximations used to derive Equation \ref{6.9} are reasonable.
Adding NaOH converts a portion of the \(\text{NH}_4^+\) to NH 3 following reaction
\[\mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(a q) \nonumber\]
Because this reaction’s equilibrium constant is so large (it is equal to ( K b ) -1 or \(5.7 \times 10^4\)), we may treat the reaction as if it goes to completion. The new concentrations of \(\text{NH}_4^+\) and NH 3 are
\[C_{\mathrm{NH}_{4}^{+}}=\frac{\operatorname{mol} \ \mathrm{NH}_{4}^{+}- \ \mathrm{mol} \mathrm{OH}^{-}}{V_{\mathrm{total}}} \nonumber\]
\[C_{\mathrm{NH}_4^+}=\frac{(0.030 \ \mathrm{M})(0.10 \ \mathrm{L})-(0.10 \ \mathrm{M})\left(1.0 \times 10^{-3} \ \mathrm{L}\right)}{0.10 \ \mathrm{L}+1.0 \times 10^{-3} \ \mathrm{L}}=0.029 \ \mathrm{M} \nonumber\]
\[C_{\mathrm{NH}_{3}}=\frac{\mathrm{mol} \ \mathrm{NH}_{3}+\mathrm{mol} \ \mathrm{OH}^{-}}{V_{\mathrm{total}}} \nonumber\]
\[C_{\mathrm{NH}_3}=\frac{(0.020 \ \mathrm{M})(0.10 \ \mathrm{L})+(0.10 \ \mathrm{M})\left(1.0 \times 10^{-3} \ \mathrm{L}\right)}{0.10 \ \mathrm{L}+1.0 \times 10^{-3} \ \mathrm{L}}=0.021 \ \mathrm{M} \nonumber\]
Substituting these concentrations into the equation 6.60 gives a pH of
\[\mathrm{pH}=9.24+\log \frac{0.021}{0.029}=9.10 \nonumber\]
Note that adding NaOH increases the pH from 9.06 to 9.10. As we expect, adding a base makes the pH more basic. Checking to see that the pH changes in the right direction is one way to catch a calculation error.
Exercise 6.8.1
Calculate the pH of a buffer that is 0.10 M in KH 2 PO 4 and 0.050 M in Na 2 HPO 4 . What is the pH after we add 5.0 mL of 0.20 M HCl to 0.10 L of this buffer. Use Appendix 11 to find the appropriate K a value.
The acid dissociation constant for \(\text{H}_2\text{PO}_4^-\) is \(6.32 \times 10^{-8}\), or a p K a of 7.199. Substituting the initial concentrations of \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) into Equation \ref{6.9} and solving gives the buffer’s pH as
\[\mathrm{pH}=7.199+\log \frac{\left[\mathrm{HPO}_{4}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]}=7.199+\log \frac{0.050}{0.10}=6.898 \approx 6.90\nonumber\]
Adding HCl converts a portion of \(\text{HPO}_4^{2-}\) to \(\text{H}_2\text{PO}_4^-\) as a result of the following reaction
\[\mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \nonumber\]
Because this reaction’s equilibrium constant is so large (it is \(1.59 \times 10^7\)), we may treat the reaction as if it goes to completion. The new concentrations of \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) are
\[C_{\mathrm{H}_{2} \mathrm{PO}_{4}^{4-}}=\frac{\mathrm{mol} \ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{mol} \ \mathrm{HCl}}{V_{\mathrm{total}}} \nonumber\]
\[C_{\mathrm{H}_{2} \mathrm{PO}_{4}^{4-}}=\frac{(0.10 \ \mathrm{M})(0.10 \ \mathrm{L})+(0.20 \ \mathrm{M})\left(5.0 \times 10^{-3} \ \mathrm{L}\right)}{0.10 \ \mathrm{L}+5.0 \times 10^{-3} \ \mathrm{L}}=0.105 \ \mathrm{M} \nonumber\]
\[C_{\mathrm{HPO}_{4}^{2-}}=\frac{\mathrm{mol} \ \mathrm{HPO}_{4}^{2-}-\mathrm{mol} \ \mathrm{HCl}}{V_{\mathrm{total}}} \nonumber\]
\[C_{\mathrm{HPO}_{4}^{2-}}=\frac{(0.05 \ \mathrm{M})(0.10 \ \mathrm{L})-(0.20 \ \mathrm{M})\left(5.0 \times 10^{-3} \ \mathrm{L}\right)}{0.10 \ \mathrm{L}+5.0 \times 10^{-3} \ \mathrm{L}}=0.0381 \ \mathrm{M} \nonumber\]
Substituting these concentrations into Equation \ref{6.9} gives a pH of
\[\mathrm{pH}=7.199+\log \frac{\left[\mathrm{HPO}_{4}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{PO}_{4}^-\right]}=7.199+\log \frac{0.0381}{0.105}=6.759 \approx 6.76 \nonumber\]
As we expect, adding HCl decreases the buffer’s pH by a small amount, dropping from 6.90 to 6.76.
We can use a multiprotic weak acid to prepare buffers at as many different pH’s as there are acidic protons, with the Henderson–Hasselbalch equation applying in each case. For example, for malonic acid (p K a1 = 2.85 and p K a2 = 5.70) we can prepare buffers with pH values of
\[\begin{array}{l}{\mathrm{pH}=2.85+\log \frac{C_{\mathrm{HM}^{-}}}{C_{\mathrm{H}_{2} \mathrm{M}}}} \\ {\mathrm{pH}=5.70+\log \frac{C_{\mathrm{M}^{2-}}}{C_{\mathrm{HM}^-}}}\end{array} \nonumber\]
where H 2 M, HM – and M 2– are malonic acid’s different acid–base forms.
Although our treatment of buffers is based on acid–base chemistry, we can extend buffers to equilibria that involve complexation or redox reactions. For example, the Nernst equation for a solution that contains Fe 2 + and Fe 3 + is similar in form to the Henderson‐Hasselbalch equation.
\[E=E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}-0.05916 \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \nonumber\]
A solution that contains similar concentrations of Fe 2 + and Fe 3 + is buffered to a potential near the standard state reduction potential for Fe 3 + . We call such solutions redox buffers. Adding a strong oxidizing agent or a strong reducing agent to a redox buffer results in a small change in potential.
Representing Buffer Solutions with Ladder Diagrams
A ladder diagram provides a simple way to visualize a solution’s predominate species as a function of solution conditions. It also provides a convenient way to show the range of solution conditions over which a buffer is effective. For example, an acid–base buffer exists when the concentrations of the weak acid and its conjugate weak base are similar. For convenience, let’s assume that an acid–base buffer exists when
\[\frac{1}{10} \leq \frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \leq \frac{10}{1} \nonumber\]
Substituting these ratios into the Henderson–Hasselbalch equation
\[\begin{aligned} \mathrm{pH} &=\mathrm{p} K_{\mathrm{a}}+\log \frac{1}{10}=\mathrm{p} K_{\mathrm{a}}-1 \\ \mathrm{pH} &=\mathrm{p} K_{\mathrm{a}}+\log \frac{10}{1}=\mathrm{p} K_{\mathrm{a}}+1 \end{aligned} \nonumber\]
shows that an acid–base buffer works over a pH range of p K a ± 1.
Using the same approach, it is easy to show that a metal‐ligand complexation buffer for ML n exists when
\[\mathrm{pL}=\log K_{n} \pm 1 \text { or } \mathrm{pL}=\log \beta_{n} \pm \frac{1}{n} \nonumber\]
where K n or \(\beta_n\) is the relevant stepwise or overall formation constant. For an oxidizing agent and its conjugate reducing agent, a redox buffer exists when
\[E=E^{\circ} \pm \frac{1}{n} \times \frac{R T}{F}=E^{\circ} \pm \frac{0.05916}{n}\left(\text { at } 25^{\circ} \mathrm{C}\right) \nonumber\]
Figure 6.8.1 shows ladder diagrams with buffer regions for several equilibrium systems.
Preparing a Buffer
Buffer capacity is the ability of a buffer to resist a change in pH when we add to it a strong acid or a strong base. A buffer’s capacity to resist a change in pH is a function of the concentrations of the weak acid and the weak base, as well as their relative proportions. The importance of the weak acid’s concentration and the weak base’s concentration is obvious. The more moles of weak acid and weak base a buffer has, the more strong base or strong acid it can neutralize without a significant change in its pH.
Although a higher concentration of buffering agents provides greater buffer capacity, there are reasons for using smaller concentrations, including the formation of unwanted precipitates and the tolerance of biological systems for high concentrations of dissolved salts.
The relative proportions of a weak acid and a weak base also affects how much the pH changes when we add a strong acid or a strong base. A buffer that is equimolar in weak acid and weak base requires a greater amount of strong acid or strong base to bring about a one unit change in pH. Consequently, a buffer is most effective against the addition of strong acids or strong bases when its pH is near the weak acid’s p K a value.
Buffer solutions are often prepared using standard “recipes” found in the chemical literature [see, for example, (a) Bower, V. E.; Bates, R. G. J. Res. Natl. Bur. Stand. (U. S.) 1955 , 55 , 197– 200; (b) Bates, R. G. Ann. N. Y. Acad. Sci. 1961 , 92 , 341–356; (c) Bates, R. G. Determination of pH , 2nd ed.; Wiley‐Interscience: New York, 1973]. In addition, there are computer programs and on‐line calculators to aid in preparing buffers [(a) Lambert, W. J. J. Chem. Educ. 1990, 67, 150–153; (b) http://www.bioinformatics.org/JaMBW/5/4/index.html .]. Perhaps the simplest way to make a buffer, however, is to prepare a solution that contains an appropriate conjugate weak acid and weak base, measure its pH, and then adjust the pH to the desired value by adding small portions of either a strong acid or a strong base.
A good “rule of thumb” when choosing a buffer is to select one whose reagents have a p K a value close to your desired pH.
[base] pH = pK a + log ––––– [acid]
pH = pK a + log [base / acid]
The K a of acetic acid is 1.77 x 10¯ 5
pK a = −log K a = −log 1.77 x 10¯ 5 = 4.752
1.00 pH = 4.752 + log ––––– 1.00 Since the log of 1 is zero, we have pH = 4.752
HAc ⇌ H + + Ac¯
1.00 pH = 4.752 + log ––––– 0.800 x = 4.752 + 0.097 = 4.849
pH = pK a + log [base / acid] x = 4.752 + log (0.800 / 1.00) x = 4.752 − 0.097 = 4.655
pH = pK a + log [base / acid] pH = 3.752 + log [0.500 / 0.700] pH = 3.752 + (−0.146) pH = 3.606
HCOOH ---> (0.700 mol/L) (0.500 L) = 0.350 mol HCOONa ---> (0.500 mol/L) (0.500 L) = 0.250 mol
NaOH ---> (1.00 mol/L) (0.0500 L) = 0.0500 mol
HCOOH ---> 0.350 mol − 0.0500 mol = 0.300 mol HCOONa ---> 0.250 mol + 0.0500 mol = 0.300 mol
pH = 3.752 + log [0.300 / 0.300] pH = 3.752 + log 1 pH = 3.752
(a) a strong acid (or base) is added to an already-buffered solution (b) a strong acid (or base) is added to a solution with just a weak base (or weak acid) in it, creating a buffer.
CH 3 NH 2 + H + ---> CH 3 NH 3 +
0.02 mol CH 3 NH 2 0.08 mol CH 3 NH 3 +
First, however, we will use the K b to get the pK a . pK a + pK b = 14 pK b = −log 5 x 10¯ 4 = 3.30103 pK a = 14 − 3.30103 = 10.69897
pH = 10.69897 + log (0.02 / 0.08) pH = 10.69897 + (−0.60206) pH = 10.09691
[H + ] = 10¯ pH = 10¯ 10.09691 = 8 x 10¯ 11 M
(0.200 mol/L) (0.0250 L) = 0.00500 mol of acetic acid (0.100 mol/L) (0.0350 L) = 0.00350 mol of NaOH
acetic acid ---> 0.00500 mol − 0.00350 mol = 0.00150 mol The acetic acid that reacts with the NaOH produces sodium acetate. In the solution will be 0.00350 mol of acetate anion (we may ignore the sodium ion. It plays no role in the pH.)
pH = 4.752 + log (0.00350 / 0.00150) pH = 4.752 + 0.368 pH = 5.120
(0.180 mol/L) (0.0500 L) = 0.00900 mol of ammonia (0.360 mol/L) (0.0050 L) = 0.00180 mol of HBr
ammonia ---> 0.00900 mol − 0.00180 mol = 0.00720 mol
NH 3 + HBr ---> NH 4 Br We can see that the ammonium ion is produced in a 1:1 molar ratio with each reactant. Since HBr is the limiting reagent, we determine that 0.00180 mole of ammonium ion will be produced. (We may ignore the bromide. It plays no role in the pH.)
pH = 9.248 + log (0.00720 / 0.00180) pH = 9.248 + 0.602 pH = 9.850
K a K b = K w (K a ) (1.77 x 10¯ 5 ) = 1.00 x 10¯ 14 K a = 5.64972 x 10¯ 10 pK a = −log 5.64972 x 10¯ 10 = 9.248
pH = pK a + log [0.25 / 0.35] The molarity of the acid (the ammonium chloride is the acid) was arrived at by dividing 0.35 mol by 1.0 L.
K w = K a K b 1.00 x 10¯ 14 = (K a ) (1.77 x 10¯ 5 ) K a = 5.65 x 10¯ 10 pK a = −log 5.65 x 10¯ 10 = 9.248 By the way, providing one constant (the K b in this example) while the solution requires a different, but related, constant (the pK a in this example) is very common in problems of this type.
pH = 9.248 + log [0.25 / 0.35] pH = 9.248 + (−0.146) = 9.10
"0.500 L of an acetic acid buffer (0.800 M total) at maximum buffering capacity" There is a lot of information in that sentence. First, since it is an "acetic acid buffer" we know that both acetic acid CH 3 COOH (or HAc) and its conjugate base acetate ion CH 3 COO¯ (or Ac¯) are present. Maximum buffering capacity means that the acid and its conjugate base are in a 1:1 molar ratio.
0.800 M (total) means 0.800 mol (total) per liter We have 0.500 L present, which means 0.400 mol (total) present Maximum buffering capacity requires that HAc and Ac¯ be present in a 1:1 molar ratio. Since 0.400 mol (total) is present, that means this is present in 0.500 L: moles HAc = 0.200 mol moles Ac¯ = 0.200 mol
moles HAc ---> 0.200 + 0.100 = 0.300 mol moles acetate ---> 0.200 − 0.100 = 0.100 mol
pH = 4.752 + log (0.100 / 0.300) pH = 4.752 + (−0.477) pH = 4.275
HAc ---> 0.400 mol salt ---> 0.400 mol
HAc ---> 0.400 mol salt ---> 0.400 + 0.100 = 0.500 mol
pH = 4.752 + log (0.500 / 0.400) pH = 4.752 + 0.097 pH = 4.849
(a) Calculate the molecular weight of the acid. (b) Calculate the K a of the acid. (c) Calculate the pOH of the original solution. (d) Calculate the pH after 40.0 mL of NaOH solution was added. (e) Calculate the percent ionization of the acid in the original solution. (f) Calculate the pH at the equivalence point.
Note that as a result of the titration TWO pieces of data emerge: (i) After we add 16 mL of the NaOH, the pH = 4.250. (ii) After we add 80 mL of the NaOH, the equivalence point is reached.
HA + OH¯ ---> H 2 O + A¯ This is commonly called a neutralization reaction.
moles = (0.500 mol/L) (0.0800 L) = 0.0400 mol
(i) the moles of strong base added (here NaOH) equals the original number of moles of acid (here HA) (ii) the titrant (here NaOH) is "chemically equivalent" to the analyte (here HA) (iii) enough NaOH has been added to "neutralize" the original HA present Based on the 1:1 molar ratio between HA and NaOH (see chemical equation above), we conclude that 0.0400 mol of acid was originally present. In other words, at the equivalence point, the moles of NaOH added equals the moles of HA present before reaction.
4.92 g / 0.0400 mol = 123 g/mol
moles = (0.500 mol/L) (0.0160 L) = 0.00800 mol
0.0400 mol − 0.00800 mol = 0.0320 mol of HA remaining. Please note that 0.0080 mol of A¯ is produced.
[base] pH = pK a + log ––––– [acid] 0.0080 4.250 = pK a + log ––––– 0.0320 Note the direct use of moles rather than molarities. 4.250 = pK a + (−0.602) pK a = 4.852 K a = 1.406 x 10¯ 5 Which, to three sig figs, is 1.41 x 10¯ 5
[H 3 O + ] [A¯] K a = –––––––––– [HA] (5.623 x 10¯ 5 ) (0.00800 + 5.623 x 10¯ 5 ) K a = ––––––––––––––––––––––––––––––– 0.0320 − 5.623 x 10¯ 5 Notice that this set up includes the amount that the HA would go down and the amount that the A¯ would go up when producing the H 3 O + in the solution. The Henderson-Hasselbalch Equation ignores amount, making the H-H an approximation. The [H 3 O + ] came from the pH of 4.250. (5.623 x 10¯ 5 ) (0.00805623) K a = –––––––––––––––––––––– 0.03194377 Ka = 1.418 x 10¯ 5 Which, to three sig figs, is 1.42 x 10¯ 5
HA + H 2 O ⇌ H 3 O + + A¯ [H 3 O + ] [A¯] K a = –––––––––– [HA] (x) (x) 1.406 x 10¯ 5 = ––––––––– 0.0800 − x Note that I used a slightly less-rounded off value for the K a . Ignoring the 'subtract x' leads to: x = (1.406 x 10¯ 5 ) (0.0800) x = 0.001060566 M pH = 2.974 pOH = 11.026
Percent ionization = the hydrogen ion concentration divided by the original acid concentration times 100 (0.0010606 M / 0.0800 M) * 100 = 1.32575% to three sig figs, 1.32%
A¯ + H 2 O ⇌ HA + OH¯ [HA] [OH¯] K b = –––––––––– [A¯]
K a K b = K w (1.406 x 10¯ 5 ) (K b ) = 1.00 x 10¯ 14 K b = 7.11238 x 10¯ 10 [A¯] = 0.0800 mol / 0.580 L = 0.137931 M Note use of the combined volumes (500 mL of weak acid mixed with 80 mL of NaOH solution.
(x) (x) 7.11238 x 10¯ 10 = –––––––– 0.137931 x = 9.90463 x 10¯ 6 M
pOH = −log 9.90463 x 10¯ 6 = 5.004 pH = 8.996
(1.80 mol/L) (0.00120 L) = 0.00216 mol
(0.850 mol/L) (0.02100 L) = 0.01785 mol
0.01785 mol − 0.00216 mol = 0.01569 mol By the way, 0.00216 mol of ammonium ion was produced by the H + reacting with the NH 3 .
pK a + pK b = pK w pK b = −log 1.77 x 10¯ 5 = 4.752 pK a + 4.752 = 14 pK a = 9.248
[base] pH = pK a + log ––––– [acid] 0.01569 pH = 9.248 + log ––––––– 0.00216 pH = 9.248 + 0.861 = 10.109
[base] pH = pK a + log ––––– [acid] 0.200 pH = 9.248 + log ––––– 0.150 pH = 9.248 + 0.125 = 9.373 Note that I used the pK a of ammonium ion.
First, you need to know the volume of buffer into which that 0.02 mol of NaOH has been added. This is because you need to know the actual moles of NH 3 and NH 4 + in the original solution, not just their concentration. So, for example, assume you have 500 mL of that original buffer. In that solution you have: (0.5 L) (0.20 mol/L) = 0.10 mol NH 3 (0.5 L) (0.15 mol/L) = 0.075 mol NH 4 +
moles NH 3 = 0.10 + 0.02 = 0.12 mol NH 3 moles NH 4 + = 0.075 − 0.02 = 0.055 mol NH 4 +
(1.00 L) (0.20 mol/L) = 0.20 mol NH 3 (1.00 L) (0.15 mol/L) = 0.15 mol NH 4 + moles NH 3 = 0.20 mol + 0.02 mol = 0.22 mol moles NH 4 + = 0.15 mol − 0.02 mol = 0.13 mol pH = 9.25 + log (0.22 / 0.13) = 9.48
MV = grams / molar mass (M) (0.250 L) = 7.45 g / 53.4916 g/mol M = 0.557097 mol/L
M 1 V 1 = M 2 V 2 (2.32 mol/L) (60.00 mL) = (x) (250.0 mL) x = 0.5568 M
0.557097 pH = 9.248 + log ––––––– 0.5568 pH = 9.248 + 0.0002316 pH = 9.248
(a) The HCl protonates some, but not all, of the NH 3 . You still have a buffer in that case. In this situation, the Henderson-Hasselbalch Equation is used. (b) There is exactly enough HCl to neutralize all of the NH 3 , leaving only NH 4 Cl in solution. That is a solution of a salt of a weak base, it is not a buffer. Do a K a caculation using the K a of NH 4 + . (c) There is excess HCl left after all the NH 3 has been protonated. In that case, you ignore all the NH 4 Cl that is in solution and treat the solution as having only a strong acid in it. The solution is not a buffer, nor is it a weak acid calculation. You are simply calculating the pH of a solution of a strong acid.
moles = MV = (0.150 mol/L) (0.500 L) = 0.0750 mol
NH 3 ---> 0.250 mol NH 4 Cl ---> 0.225 mol
NH 3 ---> 0.250 mol − 0.0750 mol = 0.175 mol NH 4 Cl ---> 0.225 mol + 0.0750 mol = 0.300 mol Note that HCl reacts with NH 3 to form NH 4 Cl in a 1:1:1 molar ratio.
pH = pK a + log ([base] / [acid]) pH = 9.248 + log (0.175 / 0.300) pH = 9.248 + (−0.234) pH = 9.014 Note that I used the pK a of ammonium, not the pK b of ammonia.
pH = 9.248 + log (0.250 / 0.225) pH = 9.248 + 0.046 pH = 9.294 Adding the HCl (an acid) has changed the pH of the buffer in the acidic direction, from 9.294 to 9.014.
moles = MV = (0.500 mol/L) (0.500 L) = 0.250 mol
NH 3 ---> 0.250 mol − 0.250 mol = 0 mol NH 4 Cl ---> 0.225 mol + 0.250 mol = 0.475 mol
NH 4 + + H 2 O ⇌ H 3 O + + NH 3
[NH 3 ] [H 3 O + ] K a = –––––––––––– [NH 4 + ]
(x) (x) 5.64937 x 10¯ 10 = –––––––– 0.158333 x = 0.0000094577 M pH = −log 0.0000094577 = 5.024 0.158333 M comes from 0.475 mol divided by 3.00 L
moles = MV = (0.650 mol/L) (0.500 L) = 0.325 mol
NH 3 ---> 0.250 mol − 0.325 mol = −0.075 mol The NH 3 is completely consumed and there is 0.075 mol of HCl left over.
0.075 mol / 3.00 L = 0.025 M
pH = −log 0.025 = 1.60 Remember, in the presence of a strong acid (HCl), the weak acid (NH 4 Cl) plays no role in determining the pH.
(a) The NaOH deprotonates some, but not all, of the NH 4 Cl. You still have a buffer in that case. In this situation, the Henderson-Hasselbalch Equation is used. (b) There is exactly enough NaOH to neutralize all of the NH 4 Cl, leaving only NH 3 in solution. That is a solution of a weak base, it is not a buffer. Do a K b caculation using the K b of NH 3 . (c) There is excess NaOH left after all the NH 3 has been deprotonated. In that case, you ignore all the NH 3 that is in solution and treat the solution as having only a strong base in it. The solution is not a buffer, nor is it a weak acid calculation. You are simply calculating the pH of a solution of a strong base.
NH 3 ---> 0.250 mol + 0.0750 mol = 0.325 mol NH 4 Cl ---> 0.225 mol − 0.0750 mol = 0.150 mol Note that NaOH reacts with NH 4 Cl to form NH 3 in a 1:1:1 molar ratio.
pH = pK a + log ([base] / [acid]) pH = 9.248 + log (0.325 / 0.150) pH = 9.248 + (0.336) pH = 9.584 Note that I used the pK a of the ammonium ion, not the pK b of ammonia.
pH = 9.248 + log (0.250 / 0.225) pH = 9.248 + 0.046 pH = 9.294 Adding the NaOH (a base) has changed the pH of the buffer in the basic direction, from 9.294 to 9.584.
moles = MV = (0.450 mol/L) (0.500 L) = 0.225 mol
NH 3 ---> 0.250 mol + 0.225 mol = 0.475 mol NH 4 Cl ---> 0.225 mol − 0.225 mol = 0 mol
NH 3 + H 2 O ⇌ NH 4 + + OH¯
[NH 4 + ] [OH¯] K b = –––––––––––– [NH 3 ]
(x) (x) 1.77 x 10¯ 5 = –––––––– 0.1583333 x = 0.00167407 M pOH = −log 0.00167407 = 2.776 pH = 14 − 2.776 = 11.224 0.1583333 M comes from 0.475 mol divided by 3.00 L
moles = MV = (0.650 mol/L) (0.500 L) = 0.350 mol
NH 4 Cl ---> 0.225 mol − 0.350 mol = −0.125 mol There is 0.125 mol of NaOH left over.
0.125 mol / 3.00 L = 0.04166667 M
pOH = −log 0.04166667 = 1.380 pH = 14 − 1.380 = 12.620 Remember, in the presence of a strong base (NaOH), the weak base (NH 3 ) plays no role in determining the pH.
(0.664 mol/L) (0.975 L) = 0.6474 mol
pH = pK a + log [base / acid] 6.420 = 4.752 + log [x / (0.6474 − x)] The x is the moles of acetate that must be present and the 0.6474 − x is the amount of acetic acid.
log [x / (0.6474 − x)] = 1.668 [x / (0.6474 − x)] = 46.5586 x = 30.142 − 46.5586x 47.5586x = 30.142 x = 0.63379 mol Acetate and KOH are in a 1:1 stoichiometric ratio, so this is the required number of moles of KOH.
0.63379 mol / 2.50 mol/L = 0.253516 L 254 mL seems like a reasonable answer
pH = 4.752 + log (0.63379 / 0.01361) = 6.420
Chemistry Steps
General Chemistry
Acids and bases.
This is a summary practice problem set on buffer solutions aimed to help identify buffers, calculating the pH of a buffer solution prepared from a weak acid and its conjugate base or vice versa.
The links to the corresponding topics are given below:
- Buffer Solutions
- The Henderson–Hasselbalch Equation
- The pH of a Buffer Solution
- Preparing a Buffer with a Specific pH
- The Common Ion Effect
- The pH and p K a Relationship
Which of the following solutions can be classified as buffer solutions?
(a) NaBr + HBr, (b) NaHSO 4 + H 2 SO 4 , (c) HCl + HOCl (d) Na 2 HPO 4 + NaH 2 PO 4 , (e) CH 3 CH 2 NH 2 + CH 3 CH 2 NH 3 + (f) NaNO 2 + HNO 2 , (g) KCN + HCN, (h) Na 2 SO 4 + NaHSO 4 , (i) NH 3 + NH 4 ClO 3 , (j) CH 3 CO 2 H + NaOH.
Addition of which salt will suppress the ionization of HOCl?
A solution of ammonia and ammonium chloride is a common buffer system. Write the corresponding reactions to show how it resists a pH change when an acid or a base is added to it.
Calculate the pH of the buffer solution consisting of 0.75 M NH 3 and 0.95 M NH 4 Cl. K b (NH 3 ) = 1.8 x 10 -5 .
Calculate the pH of the buffer solution which is 1.4 M CH 3 COONa and 1.8 M CH 3 COOH. K a (CH 3 CO 2 H) = 1.7 x 10 -5
Calculate the pH of the buffer solution that is 0.70 M NaCN and 0.55 M HCN. K a (HCN) = 4.9 x 10 -10
Calculate the pH of the buffer solution that is 0.60 M propionic acid (CH 3 CH 2 CO 2 H) and 0.75 M sodium propionate. K a (CH 3 CH 2 CO 2 H) = 1.32 x 10 -5
What is the pH of a buffer solution that is 0.85 M pyridine (C 5 H 5 N) and 1.3 M pyridinium chloride (C 5 H 5 NHCl)? K b (C 5 H 5 N) = 1.7 x 10 -9 16.82
Calculate the pH of a solution that is 0.8 M HF and 2.0 M NaF. K a (HF) = 6.6 x 10 -4
Calculate the ratio of NaNO 2 to HNO 2 required to create a buffer with pH = 4.00. K a (HNO 2 = 7.2 x 10 -4 )
Assuming no volume change, how many grams of sodium benzoate (C 7 H 5 O 2 Na) needs to be added to 250.0 mL of a 0.25 M benzoic acid (C 7 H 5 O 2 H) solution to prepare a buffer with a pH of 4.60? K a = 6.46 x 10 -5
How many mL of 0.60 M HF and 0.70 M NaF must be mixed to prepare 1.00 L of a buffer solution at pH of 4.2? p K a HF = 3.8
What is the concentration of C 6 H 5 NH 3 Cl in the buffer solution with a pH of 4.3 containing 0.50 M C 6 H 5 NH 2 ? K b (C 6 H 5 NH 2 ) 3.8 x 10 -10
Which of the following pairs is the best choice to prepare a buffer with pH = 3.2 ?
In what mass ratio would you add the components to prepare the buffer.
a) CH 3 COOH and CH 3 COONa b) HNO 2 and KNO 2 c) NH 3 and NH 4 Cl d) C 5 H 5 N and C 5 H 5 NHCl.
What would be the pH of a buffer solution composed of 0.60 M NH 3 /0.45 M NH 4 Cl after 0.010 mole of gaseous HCl is added to a 400.0 mL solution?
Calculate the pH after 20.0 g NaOH( s ) is added to 1.0 L of a buffer solution containing 1.80 M acetic acid and 1.50 M sodium acetate at pH 4.0.
- Strong Acid–Strong Base Titrations
- Titration of a Weak Acid by a Strong Base
- Titration of a Weak Base by a Strong Acid
- Titration of Polyprotic Acids
- Buffer Solutions Practice Problems
- K sp and Molar Solubility
- The Effect of a Common Ion on Solubility
- The Effect of pH on Solubility
- Will a Precipitate Form? K sp and Q
- K sp and Molar Solubility Practice Problems
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- Calculation of the pH of a Buffer Solution
- Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Strong Acid
- Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Strong Base
- Calculation of the Buffer Capacity
- First, write the equation for the ionization of acetic acid and the K a expression. Rearrange the expression to solve for the hydronium ion concentration.
- Second, determine the number of moles of acid and of the conjugate base.
- Substitute these values, along with the K a value, into the above equation and solve for the hydronium ion concentration. Convert the hydronium ion concentration into pH.
- First, write the equation for the ionization of the ammonium ion in water and the corresponding Ka expression. Rearrange the equation to solve for the hydronium ion concentration.
- Second, convert the pH back into the hydronium ion concentration and then substitute it into the above equation along with the K a . Solve for the ratio of ammonium ion to ammonia.
- First, write the equation for the ionization of acetic acid in water and the related K a expression rearranged to solve for the hydronium ion concentration.
- Second, make an "ICE" chart. Let "x" represent the hydronium ion concentration once equilibrium has been re-established. We will assume that all of the added acid is consumed.
- Substitute into the K a expression and solve for the hydronium ion concentration. Convert the answer into pH.
- First, write the equation for the ionization of the ammonium ion and the related K a expression solved for the hydronium ion concentration.
- Second, make an "ICE" chart. Let "x" be the concentration of the hydronium ion at equilibrium. The change in the amount of the ammonium ion will be equal to the amount of strong base added (075 M x 0.0200 L = 0.0015 mol).
- Third, substitute into the K a expression and solve for the hydronium ion concentration. Convert the answer into pH.
- First, write the equation for the ionization of the weak acid, in this case of hydrogen carbonate. Although this step is not truly necessary to solve the problem, it is helpful in identifying the weak acid and its conjugate base.
- Second, added strong acid will react with the conjugate base, CO 3 2- . Therefore, the maximum amount of acid that can be added will be equal to the amount of CO 3 2- , 0.50 moles.
- Third, added strong base will react with the weak acid, HCO 3 - . Therefore, the maximum amount of base that can be added will be equal to the amount of HCO 3 - , 0.35 moles.
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How do you solve chemistry buffer problems?
May 2, 2022
Buffer pH Calculations and Buffer Solutions
Buffer solutions and ph calculations.
Buffer pH calculations and problems can be quite complex, so it’s always good to have a plan — a map of “how you’re going to solve” the problem.
Let’s dive in with our first example…
ex: A buffer solution contains 0.50M acetic acid (Ka = 1.8 x 10^–5) and 0.50M sodium acetate. Calculate the pH of this solution.
➞ identify the major species first: HC2H3O2, Na+(neutral), C2H3O2- , H2O (neutral)
➞ so we have our ICE Table :
Now, because this is a [HC2H3O2 / C2H3O2-] buffer solution, or “buffer,” it resists a change in its pH upon the addition of acid (H+) or base (OH-).
For example, what would happen if we added some strong base ( i.e. NaOH) to the pH 4.74 buffer in the above problem?
ex: Calculate the change in pH that occurs when 0.010 mol NaOH(s) is added to 1.0L of the buffer prepared in the last example. Then, compare this pH change with that which occurs when 0.010 mol NaOH(s) is added to 1.0L of water.
➞ the approach:
1. realize that NaOH(s) completely dissociates and the OH- reacts completely (100%) with any acid that’s around.
- Carry out these stoichiometric calculations.
2. do the equilibrium problem (ICE Table, or ICE Method).
➞ here we go:
1. major species: Na+(neutral), OH-, HC2H3O2, C2H3O2- , H2O (neutral)
OH- is a strong base, so it will “grab” all the H+ that it can:
2. now, the equilibrium problem (after all OH- has reacted with HC2H3O2).
Wow, we added a very strong base (NaOH) to our original buffer solution, but the change in pH was minimal: 4.76–4.74 = 0.02 pH units.
This is what buffers do!
They resist large changes to their pH upon the addition of acids or bases.
But WAIT — we’re not done with the problem.
We were also asked to calculate the pH change if the strong base (NaOH) were added to pH = 7.00 water…
➞ now, if 0.010 mol NaOH had been added to 1.0L H2O instead of the buffer:
Question: How Exactly Does a Buffer Work?
Why does its pH not increase substantially when a strong base (i.e. NaOH) is added?
➞ Answer: The net result is that free OH- ions are not allowed to accumulate.
For example, OH-(aq) in the last sample problem is not around for very long, but instead it gets replaced by the much weaker base, C2H3O2-, according to:
NOTE: The same is true when a strong acid (H+) such as HCl is added to a buffer.
The strong acid (HCl) is replaced by a much weaker acid (perhaps NH4+), and the pH will only decrease by a small amount.
Next up in our discussion of SECTION 15 — Applications of Acid-Base Equilibria ,
We’ll talk about the Henderson-Hasselbalch Equation and of course, more buffers…
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Buffer Preparation
Why accurate weighing is vital for quality buffer solutions.
- Applications
Publications
Related products, faq - buffer solution preparation.
Buffer preparation is a common process in chemistry and biochemistry laboratories. A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffer solutions are used to help maintain a stable pH value of another solution that is mixed with the buffer. The buffer resists changes in the pH value of the whole solution when a small amount of a different acid or alkali is introduced into the solution either through addition or via any chemical reaction within the solution. Buffer solutions are therefore very useful in a wide variety of applications in which a relatively stable pH is required. A buffer may also be called a pH buffer, hydrogen ion buffer, or buffer solution.
For example, blood contains natural buffers to maintain a stable pH of between 7.35 and 7.45 so that our enzymes work correctly. As enzyme activity varies with pH, maintaining a constant pH is essential in biochemical assays to ensure the correct level of activity is observed. In commercial applications, buffers can be found in shampoos to prevent skin irritation, in baby lotions to inhibit the growth of bacteria, and in contact lens solutions to ensure the pH level of the fluid remains compatible with that of the eye.
Preparing buffers consists of several steps: Weighing-in the components, dissolving the components, adjusting the pH and replenishing to the final volume. As the ratio of the acid to base in a buffer is directly related to the final pH, it is vital to weigh the components with a high degree of accuracy. So, it is important that the equipment used (balance, pipettes and pH meter) are properly calibrated and have sufficient accuracy.
Watch the Video: Buffer Solution Preparation – Simple, Convenient and Accurate
Buffer solution preparation takes time and must be done with care to ensure the buffer performs as desired. When the quality of your products or your biochemical analyses depend on the performance of your buffer solutions, you want to be sure you get them right first time.
Watch the video and see how you can save time and effort preparing accurate buffer solutions with METTLER TOLEDO precision balances and pH meters.
Jump to one of the following section to explore and learn more:
- How to prepare buffer solutions? Typical procedure.
- How do buffer solutions work?
- Types of buffer solutions
- What to consider when preparing a buffer solution
- Benefits of universal buffers
- Additional tips for preparing and using buffers
- Request free support for your buffer preparation process
- The challenges of preparing a buffer solution
- METTLER TOLEDO featured solution for convenient and accurate buffer preparation
- Weighing guide: Efficient weighing workflows
- Buffer preparation – an optimized workflow with comprehensive data management
- How to ensure your pH meter is calibrated correctly
- FAQ – frequently asked questions on buffer solution preparation
How to Prepare Buffer Solutions? Typical Procedure.
Preparing buffers consists of several steps: Calculate the components (concentrations and amounts) according to the required use and target volume, weigh-in the components, dissolve the components, adjust the pH, fill-up to the final volume, label, document results and use directly or store for later usage.
- Select recipe from database
- Recalculate recipe quantities according to the required buffer volume
- Weigh compounds into the vessel
- Dissolve the compounds in a suitable solvent (typically water)
- Check and adjust the pH value by using a pH meter
- Top up the solution to the required volume
- Transfer to a storage bottle and label
- Document results
How Do Buffer Solutions Work?
Buffers are aqueous systems that resist changes in pH when small amounts of acid or base are added and are composed of a weak acid and its conjugate base. A buffer keeps the pH of a solution constant by absorbing protons that are released during reactions or by releasing protons when they are consumed by reactions. The discovery that partially neutralized solutions of weak acids or bases are resistant to changes in pH, when small amounts of strong acids or bases are added, led to the concept of the ‘buffer’.
A conjugate base is an acid that has lost a proton. HA ↔ H + + A - Acid ↔ proton + conjugate base
A conjugate acid is a base that has gained a proton. A + H + ↔ H + A Base + proton ↔ conjugate acid
An equilibrium between the dissociated and undissociated form is established. For example, weak acetic acid partially dissociates in water producing an acetate ion: CH 3 COOH ↔ H + + CH 3 COO - The undissociated acetic acid, the hydrogen ions and the dissociated ion exist in equilibrium in solution.
Sodium acetate also dissociates in water to produce the same acetate ion: CH 3 COONa ↔ Na + + CH 3 COO - The undissociated sodium acetate and the ions of sodium and acetate exist in equilibrium in solution.
An aqueous solution of a mixture of acetic acid and sodium acetate can therefore absorb H + ions from the addition of an acid through the combining of the hydrogen ions with the acetate base to produce acetic acid. Furthermore, when OH - ions are introduced into the solution through the addition of an alkali, they combine with acid molecules (H + ) to produce water. In this way, as the system attempts to restore the equilibrium, the solution resists changes to the pH value. This is how a buffer solution works.
What happens after adding acid to the buffer?
When some strong acid (more H+) is added to an equilibrium mixture of the weak acid and its conjugate base, the equilibrium is shifted to the left, in accordance with Le Chatelier’s principle.
What happens after adding a base to the buffer?
Similarly, if a strong base is added to the mixture, the hydrogen ion concentration decreases by less than the amount expected for the quantity of base added. This is because the reaction shifts to the right to accommodate for the loss of H + in the reaction with the base.
Types of Buffer Solutions
Buffer solutions consisting of a weak acid and its conjugate base are known as an acidic buffers and have a pH < 7. A buffer made with acetic acid (weak acid) and sodium acetate (conjugate base) is an acidic buffer and has a pH of around 4.75. Buffer solutions consisting of a weak base and its conjugate acid are known as alkaline buffers and have a pH > 7. An example of an alkaline buffer is an aqueous solution of ammonium hydroxide (weak base) and ammonium chloride (conjugate acid) which has a pH of 9.25.
What to Consider When Preparing a Buffer Solution
Buffer solutions work best when their pH is similar to that of the system or solution being investigated. In the study of enzymes in human biology, a system is required that matches the pH of blood, i.e. 7.35 to 7.45 otherwise the enzymes will not function correctly. If the buffer system has a pH outside of the desired range, this will also negatively impact analyses.
It is therefore necessary to be able to prepare buffer solutions of a specific pH and this can be done in different ways:
- pH adjustment The crystalline acid or base is first dissolved in a volume of water equating to around 60-70% of the final required volume of buffer solution. The pH is tested and then adjusted. If a crystalline acid is used, the pH is adjusted using a base that will not generate ions that may interfere with the system under investigation. If a crystalline base is used, the pH is adjusted using a suitable acid. Once the desired pH is achieved, the volume of the buffer solution can be topped up with water to obtain the required volume.
- Mixing acid and base solutions In this method, a solution of the acid or base is mixed with a solution of the associated salt. The concentrations of the source solutions must be the same as the required buffer solution. The solutions can be mixed in varying ratios to achieve differing pH values in the final buffer solution. Alternatively, the pH can be monitored while adding one solution to the other.
- Using Henderson-Hasselbach equation The Henderson-Hasselbach equation can be used to estimate the pH of a buffer solution using the dissociation constant, pK a . When a weak acid (HA) is in solution and the dissociation into hydrogen ions (H + ) and its conjugate base (A - ) is in equilibrium, the dissociation constant is a measure of the strength of the acid at this equilibrium point. The Henderson-Hasselbach equation is as follows:
Where: pKa is the dissociation constant of the weak acid [A-] is the concentration of the conjugate base at equilibrium [HA] is the concentration of the acid at equilibrium In the case where the concentration at equilibrium of the conjugate base and the acid are the same, the pH becomes equal to the dissociation constant. At this point, the buffer solution provides the maximum buffering capability. The Henderson-Hasselbach equation is also used for determining the dissociation constant of weak acids via direct measurement of pH
Benefits of Universal Buffers Universal buffer solutions consist of a combination of several acid-base pairs. This enables universal buffers to maintain the pH value of a solution across a wider range and can therefore be used in many more applications.
Additional Tips for Preparing and Using Buffers
- Create a buffer preparation SOP It is a good idea to document the buffer preparation process and be sure that everyone follows the same procedure to ensure consistency and reproducibility. The SOP should include details of the materials used and the precise steps of when and how to add the components and measure the pH. The SOP may also include many of the points mentioned below.
- Wear protective clothing Appropriate personal protective equipment (PPE), for example protective eyewear and clothing must be worn, especially when working with strong acids or bases.
- Check for microbial contamination (especially in biological applications) Before using any buffer, check the container for microbial contamination. Buffer solutions with a pH close to neutral are particularly susceptible to contamination. You may see some cloudiness in the solution or contaminants that have settled on the bottom.
- Use your pH meter correctly Your pH meter should be regularly calibrated and maintained to ensure accurate pH measurements. The electrode should be properly prepared prior to use and sufficient buffer solution should be used to ensure that the electrode junction is correctly submerged. Be sure to wait for a stable pH value before taking the reading and rinse the electrode with distilled water afterwards. Make sure to use your pH meter at the ambient temperature or use an electrode with integrated temperature probe.
- Be aware of the temperature The amount of dissociation can vary with temperature. Your buffer solution should therefore be prepared at the same temperature at which you will perform your analyses. Also make sure that the temperature at which you calibrate the electrode is the same as the temperature at which you measure.
- Be aware of the concentration It is common practice to dilute stock buffer solutions to the required concentration for use in analyses. However, changing the concentration can affect the amount of dissociation. As pH is a measure of hydrogen ions (H + ), a change in dissociation can cause a change in pH. After dilution pH should be rechecked prior to using the buffer.
Do You Need Support?
Buffer solutions are vitally important in a wide range of applications. However, buffer preparation is time-consuming and labor-intensive and must be done with care to avoid errors. If you need support with buffer preparation or want advice on which balance meets your accuracy requirements, our team of experts is here for you. Do not hesitate to contact us for help or advice.
The Challenges of Preparing a Buffer Solution
Whilst the simplest buffer solution may consist just of an acid, a base and water, laboratory technicians often have to prepare several buffer solutions per day – typically consisting of 2-5 components, but could be as many as 20. A typical laboratory may have more than 20 buffer solution recipes that are intended to produce 1 liter of buffer.
Buffer solution calculations
With a database of 20+ recipes, the laboratory technician must first be sure that the correct buffer solution recipe is selected. For buffer solution volumes other than 1 liter, it is necessary to recalculate the quantities of all the components accordingly and log the new amounts. Any errors in the calculations or in recording the new values may lead to incorrect pH values in the buffer solution. There is a higher risk of errors in a manual data recording system.
Weighing and logging weight results
When weighing in all the different buffer solution components, care must be taken to use the right amount of the right component. The actual weight of each component must be logged, and, if recording results by hand or entering the values into a computer, care must be taken to avoid transcription errors.
Depending on how much of each of the buffer solution components is required and the minimum weight of the balance, it may be necessary to use two different balances. This complicates the process and reduces efficiency.
Correct performance of pH meters
Checking the pH level of the buffer solution is of vital importance. However, if the pH meter used is not calibrated and maintained correctly, pH readings may be inaccurate. Using a buffer solution with an incorrect pH may seriously impact subsequent analyses or product quality.
Documentation
The final buffer solution must be carefully labelled with all information to avoid any mix-ups. The expiry date is important to ensure the buffer solution is still effective when it is used. All data relating to the buffer solution preparation must be logged and stored securely for future reference and traceability.
METTLER TOLEDO Featured Solution for Convenient and Accurate Buffer Preparation
METTLER TOLEDO's XSR precision balances simplify buffer solution preparation and significantly reduce the workload for laboratory technicians.
Saved recipes
Up to 50 buffer solution recipes can be saved directly on the balance as weighing methods. By saving the most commonly used buffer solution recipes on the balance, it is quick to retrieve the method needed and start working. The balance provides step-by-step instructions so analysts can easily keep track of where they are in the buffer preparation process.
Automatic calculations and weighing-in guide
At the start of each method, the required volume of the buffer solution is entered and the component quantities are calculated automatically. As each ingredient is weighed in, the SmartTrac weighing guide displays the actual weight against the target weight graphically. The colored bar turns green as soon as the weight of the added component is within the predefined tolerance range. This enables analysts to weigh in more quickly and with greater certainty.
Efficient processes and automatic data recording
After each component has been added, the balance saves the actual weight result and then tares automatically, assisting in the weighing process and eliminating the need to manually record the results. The balance can be set to print out the weight results automatically once all the ingredients have been added.
Convenient pH measurement
METTLER TOLEDO's SevenExcellence pH meters are the perfect choice for accurate pH measurements. These easy-to-use instruments offer touchscreen operation, direct measurements at the touch of a button, and intelligent sensor management.
pH measurements are only as accurate as the calibration solutions used. METTLER TOLEDO also offers a selection of quality pH calibration solutions to match your specific requirements.
Guide: Biological Buffer Preparation
Discover how you can make your buffer preparation process easy and reliable.
Download the guide and get useful tips and hints to help improve the productivity of your preparation process, ensure quality standards and traceability for buffer solutions.
- Step-by-step SOP user guidance and accessible buffer recipes directly on the balance screen improves workflows
- Accurate weighing in compounds is vital for the quality of buffer solutions
- Exact and reliable pH measurement is important for the correct preparation of buffers
- Using and operating pipettes the correct way helps to ensure that pH buffer is properly adjusted
- Automatic calculation, labeling and reporting ensure safe data transfer and data traceability
- Download the free guide now!
Buffer Preparation – An Optimized Workflow with Comprehensive Data Management
METTLER TOLEDO Excellence Line precision balances (XPR and XSR) and SevenExcellence pH meters can be connected to our proprietary LabX laboratory software to create a fully optimized system that provides a range of key benefits:
- Automatic calculations and documentation
- All results saved in a secure database
- Full traceability
- Central management of tasks, users and instruments
Support with FDA 21CFR part 11 compliance
LabX provides comprehensive data management functions, practically eliminating manual data transcription – even detailed labels can be printed automatically. With a centralized database, all connected users and instruments can access the same stored SOPs for buffer preparation. This ensures that every analyst carries out buffer preparation with identical settings and parameters. All results and process metadata are saved automatically in a secure database so full traceability is assured.
How to Ensure Your pH Meter is Calibrated Correctly
To accurately calibrate your pH meter, you need accurate calibration buffer solutions of known pH! METTLER TOLEDO offers a range of quality pH buffers and guarantees maximum precision with NIST/DIN buffers.
Each buffer is supplied with a test certificate to help you ensure compliance and traceability. Whether you just need a technical buffer or a buffer certified by an accredited body, we have the right buffers to meet your needs.
Precision Balances and Scales
LabX Laboratory Software
SevenExcellence pH Meter Line
Laboratory pH Buffer
Faq – frequently asked questions on buffer solution preparation.
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1. How can I make weighing for buffer preparation more efficient?
On METTLER TOLEDO XPR/XSR balances, you can save your own specific weighing methods. For buffer preparation, this is achieved by using the 'Simple formulation with templates' method on the balance to store each buffer component, with its target weight and tolerance interval, as a recipe on the balance. When the method is started, you are guided through each step of the procedure. No time is wasted checking a recipe sheet and there is no uncertainty about where you are in the procedure. Weight results are saved automatically and can be printed out at the end so you save a considerable amount of time by eliminating the need to record all the weight results by hand.
2. Which is the best balance for buffer solution preparation?
The best balance depends on your individual requirements. You need to know the largest and the smallest amounts you want to weigh as well as how accurately you need to weigh them (i.e. what the tolerance or allowable error is). METTLER TOLEDO offers a free service to help you choose the right balance to meet your application requirements. Simply ask your local representative for a free balance recommendation. The same service also determines if your existing balance meets your requirements.
- GWP Recommendation
3. During buffer solution preparation, there's such a lot of data to record. How can I avoid making mistakes? I need an error-free solution.
First, connect a barcode reader to your balance. This will enable you to electronically record sample IDs, lot numbers, and order numbers etc. Second, connect a printer to your balance. With the METTLER TOLEDO P-5x line of printers , the metadata and the date and time can be automatically printed out along with the results at the end of the weighing procedure. An alternative option is to connect your balance to LabX laboratory software which offers comprehensive data management including customizable reports that can be sent straight to your LIMS or ERP.
4. How can I be sure of the pH reading of my buffer solution?
Periodic calibration and regular checks of your pH meter are necessary in order to be sure of its accuracy. Consult the guide and learn more about METTER TOLEDO's solution for easy calibration of a pH meter . And check out this video: pH Calibration ‒ A Handy How-To Guide.
5. What if I add more of one ingredient than is needed?
If you accidentally weigh in too much of one ingredient, it is possible to add additional amounts of the other ingredients rather than discard the materials weighed out up to that point. If calculating the amounts manually, this can be challenging and there is a high risk of errors. You should also be aware that the additional amounts required may be very small and your balance may not be suitable for this. A second balance with a higher degree of accuracy and a lower minimum weight may be required. If your balance is connected to LabX, LabX can perform the necessary recalculations for you and keep track of your progress in the procedure.
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So let's do that. Get some more space down here. So the pH is equal to the pKa, which again we've already calculated in the first problem is 9.25 plus
This chemistry video tutorial explains how to calculate the pH of a buffer solution using the henderson hasselbalch equation.
Perhaps the simplest way to make a buffer, however, is to prepare a solution that contains an appropriate conjugate weak acid and weak base
What is its pH? Solution: 1) To solve the above example, we must know the pKa of acetic acid. Often, the problem will provide
This is a summary practice problem set on buffer solutions aimed to help identify buffers, calculating the pH of a buffer solution prepared from a weak acid
New type of problem to solve! – solutions with two things dissolved. Example: Calculate [H3O+] in a solution that is 0.10 M in HF and 0.20 M in NaF. Also.
In order to calculate the pH of the buffer solution you need to know the amount of acid and the amount of the conjugate base combined to make the solution.
pH=pK_a + log_10{[[A^-]]/[[HA])} See this old answer.. As you have probably heard ad nauseaum, at the point of half equivalence in the
Buffer pH calculations and problems can be quite complex, so it's always good to have a plan — a map of “how you're going to solve” the problem.
The Challenges of Preparing a Buffer Solution ... Whilst the simplest buffer solution may consist just of an acid, a base and water, laboratory technicians often