Question 17.12: Calculating the pH of a Buffer When a Strong Acid or Strong ...
Calculating the pH of a Buffer When a Strong Acid or Strong Base Is Added
Calculate the \mathrm{pH} of 75 \mathrm{~mL} of the buffer solution described in Example 17.10 (0.10 M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} and 0.20 M \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} ) to which 9.5 \mathrm{~mL} of 0.10 M hydrochloric acid is added. Compare the \mathrm{pH} change with what would occur if this amount of acid were added to pure water.
PROBLEM STRATEGY
Do the problem in two parts. First, you assume that the \mathrm{H}_{3} \mathrm{O}^{+}ion from the strong acid and the conjugate base from the buffer react completely. This is a stoichiometric calculation. Actually, the \mathrm{H}_{3} \mathrm{O}^{+}ion and the base from the buffer reach equilibrium just before complete reaction. So you now solve the equilib- rium problem using concentrations from the stoichiometric calculation. Because these concentrations are not far from equilibrium, you can use the usual simplifying assumption about x.
When hydronium ion (from hydrochloric acid) is added to the buffer, it reacts with acetate ion.
\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q) \longrightarrow \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
Because acetic acid is a weak acid, you can assume as a first approximation that the reaction goes to completion. This part of the problem is simply a stoichiometric calculation. Then you assume that the acetic acid ionizes slightly. This part of the problem involves an acid-ionization equilibrium.
STOICHIOMETRIC CALCULATION You must first calculate the amounts of hydrogen ion, acetate ion, and acetic acid present in the solution before reaction. The molar amount of hydrogen ion will equal the molar amount of hydrochloric acid added, which you obtain by converting the volume of hydrochloric acid, \mathrm{HCl}, to moles of \mathrm{HCl}. To do this, you note for 0.10 \mathrm{M} \mathrm{HCl} that 1 \mathrm{~L} \mathrm{HCl} is equivalent to 0.10 \mathrm{~mol} \mathrm{HCl}. Hence, to convert 9.5 \mathrm{~mL}\left(=9.5 \times 10^{-3} \mathrm{~L}\right) of hydrochloric acid, you have
9.5 \times 10^{-3} \cancel{\mathrm{LHCl} } \times \frac{0.10 \mathrm{~mol } \mathrm{ HCl}}{1 \cancel{\mathrm{LHCl} }}=0.00095 \mathrm{~mol} \mathrm{HCl}
Hydrochloric acid is a strong acid, so it exists in solution as the ions \mathrm{H}_{3} \mathrm{O}^{+}and \mathrm{Cl}^{-}. Therefore, the amount of \mathrm{H}_{3} \mathrm{O}^{+}added is 0.00095 \mathrm{~mol}.
The amounts of acetate ion and acetic acid in the 75-\mathrm{mL} sample are found in a similar way. The buffer in Example 17.10 contains 0.20 \mathrm{~mol} of acetate ion and 0.10 \mathrm{~mol} of acetic acid in 1 \mathrm{~L} of solution. The amounts in 75 \mathrm{~mL}(=0.075 \mathrm{~L}) of solution are obtained by converting to moles.
\begin{aligned} & 0.075 \cancel{\text { L soln }} \times \frac{0.20 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}}{1 \cancel{\text { L soln }}}=0.015 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-} \\ & 0.075 \text { Lsotn } \times \frac{0.10 \mathrm{~mol} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}}{1 \cancel{\text { L soln }}}=0.0075 \mathrm{~mol} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \end{aligned}
You now assume that all of the hydronium ion added (0.00095 mol) reacts with acetate ion. Therefore, 0.00095 \mathrm{~mol} of acetic acid is produced and 0.00095 mol of acetate ion is used up. Hence, after reaction you have
\text { Moles of acetate ion }=(0.015-0.00095) \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}=0.014 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}Moles of acetic acid =(0.0075+0.00095) mol \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}=0.0085 \mathrm{~mol} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}
EQUILIBRIUM CALCULATION You first calculate the concentrations of \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} and \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}present in the solution before you consider the acid-ionization equilibrium. Note that the total volume of solution (buffer plus hydrochloric acid) is 75 \mathrm{~mL}+9.5 \mathrm{~mL}, or 85 \mathrm{~mL}(0.085 \mathrm{~L}). Hence, the starting concentrations are
\begin{gathered} {\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=\frac{0.0085 \mathrm{~mol}}{0.085 \mathrm{~L}}=0.10 \mathrm{M}} \\ {\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}\right]=\frac{0.014 \mathrm{~mol}}{0.085 \mathrm{~L}}=0.16 \mathrm{M}} \end{gathered}
From this, you construct the following table:
\begin{array}{cccc} \text { Concentration }(M) & \mathrm{HC}_{2} \mathbf{H}_{3} \mathbf{O}_{2}(\mathbf{a q})+\mathrm{H}_{2} \mathbf{O}(l)& \rightleftharpoons \mathrm{H}_{3} \mathbf{O}^{+}(a q)&+\mathrm{C}_{2} \mathrm{H}_{3} \mathbf{O}_{2}{ }^{-}(a q) \\ \text { Starting } & 0.10 & \sim 0 & 0.16 \\ \text { Change } & -x & +x & +x \\ \text { Equilibrium } & 0.10-x & x & 0.16+x \end{array}
The equilibrium-constant equation is
\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\right]}{\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]}=K_{a}
Substituting, you get
\frac{x(0.16+x)}{0.10-x}=1.7 \times 10^{-5}
If you assume that x is small enough that 0.16+x \simeq 0.16 and 0.10-x \simeq 0.10, this equation becomes
\frac{x(0.16)}{0.10}=1.7 \times 10^{-5}
or
x=1.7 \times 10^{-5} \times \frac{0.10}{0.16}=1.1 \times 10^{-5}
Note that x is indeed small, so the assumptions you made earlier are correct. The \mathrm{H}_{3} \mathrm{O}^{+}concentration is 1.0 \times 10^{-5} \mathrm{M}. The \mathrm{pH} is
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left(1.1 \times 10^{-5}\right)=\mathbf{4 . 9 6}
Because the \mathrm{pH} of the buffer was 5.07 (see Example 17.10), the \mathrm{pH} has changed by 5.07-4.96=0.11 units.
ADDING HCl TO PURE WATER If 9.5 \mathrm{~mL} of 0.10 \mathrm{M} hydrochloric acid were added to 75 \mathrm{~mL} of pure water, the hydronium-ion concentration would change to
\begin{aligned} {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] } & =\frac{\text { amount of } \mathrm{H}_{3} \mathrm{O}^{+} \text {added }}{\text { total volume of solution }} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] } & =\frac{0.00095 \mathrm{~mol} \mathrm{H}_{3} \mathrm{O}^{+}}{0.085 \mathrm{~L} \text { solution }}=0.011 \mathrm{M} \end{aligned}
(The total volume is 75 \mathrm{~mL} of water plus 9.5 \mathrm{~mL} \mathrm{HCl}, assuming no change of volume on mixing.) The \mathrm{pH} is
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log (0.011)=1.96
The \mathrm{pH} of pure water is 7.00 , so the change in \mathrm{pH} is 7.00-1.96=5.04 units, compared with 0.11 units for the buffered solution.