Question 16.4: Calculate the pH of 1.00 L of a 0.10 M acetic acid–0.10 M so......
Calculate the pH of 1.00 L of a 0.10 M acetic acid–0.10 M sodium acetate solution with an initial pH of 4.74 after the addition of
(a) 0.01 mol of OH^{-} (b) 0.01 mol of H_{3}O^{+}
STRATEGY
Because neutralization reactions involving strong acids or strong bases go essentially 100% to completion (Section 16.1), we must write the reaction and set up an equilibrium table to find [H_{3}O^{+}].
| IDENTIFY | |
| Known | Unknown |
| Volume of buffer solution (1.00 L) | Final pH |
| Initial concentration of acetic acid and acetate ion (0.10 M) | |
| Amount of acid or base added (0.01 mol) | |
(a)
When 0.01 mol of OH^{-} is added to the buffer solution, it will react with the weak acid (CH_{3}CO_{2}H). The neutralization reaction will alter the numbers of moles. We set up an equilibrium table to keep track of the change, putting the initial number of moles of acetic acid and acetate ion in the first row. Initially, we have (1.00 L)(0.10 mol/L) = 0.10 mol of acetic acid and an equal amount of acetate ion.
Table 1
If we assume that the solution volume remains constant at 1.00 L, the concentrations of the buffer components after neutralization are
[CH_{3}CO_{2}H]=\frac{0.09 mol}{1.00 L}=0.09 M[CH_{3}CO_{2} ^{-}]=\frac{0.11 mol}{1.00 L}=0.11 M
Substituting these concentrations into the expression for [H_{3}O^{+}], we can then calculate the pH:
[H_{3}O^{+}]=K_{a}\frac{[CH_{3}CO_{2}H]}{[CH_{3}CO_{2} ^{-}]}=(1.8\times 10^{-5})(\frac{0.09}{0.11})=1.5\times 10^{-5} M
pH = 4.82
(b)
When 0.01 mol of H_{3}O^{+} is added to the buffer solution, it will react with the weak base (CH_{3}CO_{2} ^{-}). The added strong acid will convert 0.01 mol of acetate ions to 0.01 mol of acetic acid because of the neutralization reaction.
Table 2
The concentrations after neutralization will be [CH_{3}CO_{2}H] = 0.11 M and [CH_{3}CO_{2} ^{-}] = 0.09 M, and the pH of the solution will be 4.66:
[H_{3}O^{+}]=K_{a}\frac{[CH_{3}CO_{2}H]}{[CH_{3}CO_{2} ^{-}]}=(1.8\times 10^{-5})(\frac{0.11}{0.09})=2.2\times 10^{-5} M
pH = 4.66
Table 1
| Neutralization Reaction | CH_{3}CO_{2}H(aq) | + | OH^{-}(aq) | \overset{100\%}{\longrightarrow} | H_{2}O(l) | + | CH_{3}CO_{2} ^{-}(aq) |
| Before reaction (mol) | 0.10 | 0.01 | 0.10 | ||||
| Change (mol) | – 0.01 | – 0.01 | + 0.01 | ||||
| After reaction (mol) | 0.09 | ∼ 0 | 0.11 |
Table 2
| Neutralization Reaction | CH_{3}CO_{2} ^{-}(aq) | + | H_{3}O^{+}(aq) | \overset{100\%}{\longrightarrow} | H_{2}O(l) | + | CH_{3}CO_{2}H(aq) |
| Before reaction (mol) | 0.10 | 0.01 | 0.10 | ||||
| Change (mol) | – 0.01 | – 0.01 | + 0.01 | ||||
| After reaction (mol) | 0.09 | ∼ 0 | 0.11 |