Example 13.20: pH Changes in Buffered and Unbuffered Solutions

(a) Calculate the pH of an acetate buffer that is a mixture with 0.100 M acetic acid and 0.100 M sodium acetate.

Solution

To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):

Determine the direction of change. The equilibrium in a mixture of \(\ce{H3O^+}\), \(\ce{CH3CO2-}\), and \(\ce{CH3CO2H}\) is:

\(\ce{CH3CO2H(aq)}\)\(\ce{ + }\)\(\ce{H2O(l)}\)\(\ce{<=>}\)\(\ce{H3O+(aq)}\)\(\ce{ + }\)\(\ce{CH3CO2-(aq)}\)\(\ce{ }\)

The equilibrium constant for \(\ce{CH3CO2H}\) is not given, so we look it up in Appendix H:

\(K_{\mathrm{a}}\) \(= 1.8\times 10^{-5}\)


\([\ce{CH3CO2H}]_{\mathrm{init}}\) \(= 0.10\)


\([\ce{CH3CO2-}]_{\mathrm{init}}\) \(= 0.10\)


With

\([\ce{CH3CO2H}]_{\mathrm{init}} = [\ce{CH3CO2-}]_{\mathrm{init}} = 0.10\ \mathrm{M} \cdot \mathrm{and} \cdot [\ce{H3O+}] = ~_{\mathrm{0}} \cdot 1\ \mathrm{M}\)     

the reaction shifts to the right to form \(\ce{H3O+}\).
Determine x and equilibrium concentrations. A table of changes and concentrations follows:

Solve for x and the equilibrium concentrations. We find:

\(K_{\mathrm{a}} = \dfrac{[\ce{CH3CO2-}]_{\mathrm{eq}} \cdot [\ce{H3O+}]}{[\ce{CH3CO2H}]_{\mathrm{eq}}} = \dfrac{([\ce{CH3CO2-}]_{\mathrm{init}} + x) \cdot x}{[\ce{CH3CO2H}]_{\mathrm{init}} - x}\)     

\(x = \dfrac{K_{\mathrm{a}} \cdot ([\ce{CH3CO2H}]_{\mathrm{init}} - x)}{[\ce{CH3CO2-}]_{\mathrm{init}} + x}\)     

\(x\) \(= K_{\mathrm{a}}\)

\(\ \ \ =1.8\times 10^{-5}\)

\(\ \ \ =1.8\times 10^{-5}\)


and

\([\ce{H3O+}]\) \(= 0 + x\)

\(\ \ \ =0 + 1.8\times 10^{-5}\)

\(\ \ \ =1.8\times 10^{-5}\)


Thus:

\(\mathrm{pH}\) \(= -\mathrm{log}([\ce{H3O+}])\)

\(\ \ \ =-\mathrm{log}(1.8\times 10^{-5})\)

\(\ \ \ =4.74\)


Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction \(Failed to interpret math QQQcoefficient, Q QQQ = K_{\mathrm{a}}\).


(b) Calculate the pH after 1.00 mL of 0.100 M \(\ce{NaOH}\) is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.
First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:

Determine the chemical amount of \(\ce{NaOH}\). One milliliter (0.0010 L) of 0.10 M \(\ce{NaOH}\) contains:

\(c_{\mathrm{\ce{NaOH}}}\) \(= 0.100\ \mathrm{M}\)


\(V_{\mathrm{\ce{NaOH}}}\) \(= 1.00\ \mathrm{mL}\)


\(n_{\mathrm{\ce{NaOH}}}\) \(= c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH}}}\)

\(\ \ \ =0.100\ \mathrm{M} \cdot 1.00\ \mathrm{mL}\)

\(\ \ \ =1.00\times 10^{-4}\ \mathrm{mol}\)


Determine the chemical amount of \(\ce{CH2CO2H}\). Before reaction, 100 mL of the buffer solution contains:

\(V_{\mathrm{\ce{CH3CO2H}}}\) \(= 100\ \mathrm{mL}\)


\(c_{\mathrm{\ce{CH3CO2H}}}\) \(= 0.100\ \mathrm{M}\)


\(n_{\mathrm{\ce{CH3CO2H}}}\) \(= V_{\mathrm{\ce{CH3CO2H}}} \cdot c_{\mathrm{\ce{CH3CO2H}}}\)

\(\ \ \ =100\ \mathrm{mL} \cdot 0.100\ \mathrm{M}\)

\(\ \ \ =0.0100\ \mathrm{mol}\)


\(n_{\mathrm{\ce{NaCH3CO2}}}\) \(= n_{\mathrm{\ce{CH3CO2H}}}\)    (same volume and concentration as \(\ce{CH3CO2H}\))

\(\ \ \ =0.0100\ \mathrm{mol}\)

\(\ \ \ =0.0100\ \mathrm{mol}\)


Solve for the amount of \(\ce{NaCH3CO2}\) produced. The 1.0 × 10^-4 mol of \(\ce{NaOH}\) neutralizes 1.0 × 10^-4 mol of \(\ce{CH3CO2H}\), leaving:

\(n_{\mathrm{\ce{CH3CO2H},mix}}\) \(= n_{\mathrm{\ce{CH3CO2H}}} - n_{\mathrm{\ce{NaOH}}}\)

\(\ \ \ =0.0100\ \mathrm{mol} - 1.00\times 10^{-4}\ \mathrm{mol}\)

\(\ \ \ =9.9\times 10^{-3}\ \mathrm{mol}\)


and producing 1.0 × 10^-4 mol of \(\ce{NaCH3CO2}\). This makes a total of:

\(n_{\mathrm{\ce{NaCH3CO2},mix}}\) \(= n_{\mathrm{\ce{NaCH3CO2}}} + n_{\mathrm{\ce{NaOH}}}\)

\(\ \ \ =0.0100\ \mathrm{mol} + 1.00\times 10^{-4}\ \mathrm{mol}\)

\(\ \ \ =0.0101\ \mathrm{mol}\)


Find the concentration of the products. After reaction, \(\ce{CH3CO2H}\) and \(\ce{NaCH3CO2}\) are contained in 101 mL of the intermediate solution, so:

\(V_{\mathrm{mix}}\) \(= 101.\ \mathrm{mL}\)


\([\ce{CH3CO2H}]_{\mathrm{mix}}\) \(= \dfrac{n_{\mathrm{\ce{CH3CO2H},mix}}}{V_{\mathrm{mix}}}\)

\(\ \ \ =\dfrac{9.9\times 10^{-3}\ \mathrm{mol}}{101.\ \mathrm{mL}}\)

\(\ \ \ =0.098\ \mathrm{M}\)


\([\ce{NaCH3CO2}]_{\mathrm{mix}}\) \(= \dfrac{n_{\mathrm{\ce{NaCH3CO2},mix}}}{V_{\mathrm{mix}}}\)

\(\ \ \ =\dfrac{0.0101\ \mathrm{mol}}{101.\ \mathrm{mL}}\)

\(\ \ \ =0.100\ \mathrm{M}\)


Now we calculate the pH after the intermediate solution, which is 0.098 M in \(\ce{CH3CO2H}\) and 0.100 M in \(\ce{NaCH3CO2}\), comes to equilibrium. The calculation is very similar to that in part (a) of this example:

This series of calculations gives a pH of 4.75. Thus the addition of the base barely changes the pH of the solution ( Figure 17).


(c) For comparison, calculate the pH after 1.00 mL of 0.100 M \(\ce{NaOH}\) is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a \(1.8\times 10^{-5 }\)M solution of \(\ce{HCl}\)). The volume of the final solution is 101 mL.

Solution

\(V_{\mathrm{\ce{HCl}}}\) \(= 100.\ \mathrm{mL}\)


\(c_{\mathrm{\ce{HCl}}}\) \(= 1.8\times 10^{-5}\ \mathrm{M}\)


This 1.8 × 10^-5-M solution of \(\ce{HCl}\) has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:

\(n_{\mathrm{\ce{HCl}}}\) \(= c_{\mathrm{\ce{HCl}}} \cdot V_{\mathrm{\ce{HCl}}}\)

\(\ \ \ =1.8\times 10^{-5}\ \mathrm{M} \cdot 100.\ \mathrm{mL}\)

\(\ \ \ =1.8\times 10^{-6}\ \mathrm{mol}\)


As shown in part (b), 1 mL of 0.10 M \(\ce{NaOH}\) contains 1.0 × 10^-4 mol of \(\ce{NaOH}\). When the \(\ce{NaOH}\) and \(\ce{HCl}\) solutions are mixed, the \(\ce{HCl}\) is the limiting reagent in the reaction. All of the \(\ce{HCl}\) reacts, and the amount of \(\ce{NaOH}\) that remains is:

\(n_{\mathrm{\ce{NaOH},mix}}\) \(= n_{\mathrm{\ce{NaOH}}} - n_{\mathrm{\ce{HCl}}}\)

\(\ \ \ =1.00\times 10^{-4}\ \mathrm{mol} - 1.8\times 10^{-6}\ \mathrm{mol}\)

\(\ \ \ =9.8\times 10^{-5}\ \mathrm{mol}\)


The concentration of \(\ce{NaOH}\) is:

\(c_{\mathrm{\ce{NaOH},mix}}\) \(= \dfrac{n_{\mathrm{\ce{NaOH},mix}}}{V_{\mathrm{mix}}}\)

\(\ \ \ =\dfrac{9.8\times 10^{-5}\ \mathrm{mol}}{101.\ \mathrm{mL}}\)

\(\ \ \ =9.7\times 10^{-4}\ \mathrm{M}\)


The \(\ce{pOH}\) of this solution is:

\(\mathrm{pOH}\) \(= -\mathrm{log}(\dfrac{c_{\mathrm{\ce{NaOH},mix}}}{1\ \mathrm{M}})\)

\(\ \ \ =-\mathrm{log}(\dfrac{9.7\times 10^{-4}\ \mathrm{M}}{1\ \mathrm{M}})\)

\(\ \ \ =-\mathrm{log}(9.72\times 10^{-4})\)

\(\ \ \ =3.01\)


The pH is:

\(\mathrm{pH}\) \(= 14.00 - \mathrm{pOH}\)

\(\ \ \ =14.00 - 3.01\)

\(\ \ \ =10.99\)

Warning: Updated value of \(\mathrm{pH}\)


The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of \(\ce{NaOH}\) was added to the buffered solution described in part (b).