Question 16.9: Assume that you are titrating 30.0 mL of a 0.0600 M solution......
Assume that you are titrating 30.0 mL of a 0.0600 M solution of the protonated form of the amino acid methionine (H_{2}A^{+}) with 0.0900 M NaOH. Calculate the pH after addition of the following volumes of 0.0900 M NaOH:
(a) 20.0 mL (b) 30.0 mL (c) 35.0 mL
Methionine cation (H_{2}A^{+})
K_{a1} = 5.2\times 10^{–3}
K_{a2} = 6.2\times 10^{–10}
STRATEGY
First calculate the number of millimoles of H_{2}A^{+} present initially and the number of millimoles of NaOH added. Then you can tell which species remain after neutralization and therefore what type of equilibrium problem you must solve.
| IDENTIFY | |
| Known | Unknown |
| Volume and molarity of H_{2}A^{+} (30.0 mL, 0.0600 M) | pH at various points in the titration |
| Molarity of NaOH (0.0900 M) | |
| Volume of NaOH (20.0 mL, 30.0 mL, and 35.0 mL) | |
(a) The number of millimoles of H_{2}A^{+} present initially and the number of millimoles NaOH added are
mmol H_{2}A^{+} initial = (30.0 mL)(0.0600 mmol/mL) = 1.80 mmol
mmol NaOH added = (20.0 mL)(0.0900 mmol/mL) = 1.80 mmol
The added base is just enough to reach the first equivalence point, converting all the H_{2}A^{+} to HA because of the neutralization reaction.
Table 1
Therefore, the pH equals the average of pK_{a1} and pK_{a2}:
pH=\frac{pK_{a1} + pK_{a2}}{2}=\frac{[-log (5.2\times 10^{-3})]+[-log (6.2\times 10^{-10})]}{2}=\frac{2.28+9.21}{2}=5.74(b) The amount of NaOH added to the 1.80 mmol of H_{2}A^{+} present initially is (30.0 mL) × (0.0900 mmol/mL) = 2.70 mmol. The added base is enough to convert all of the H_{2}A^{+} to 1.80 mmol of HA and then convert 2.70 – 1.80 = 0.90 mmol of the resulting HA to 0.90 mmol of A^{-} because of the neutralization reaction.
Table 2
After neutralization, we have an HA–A^{-} buffer solution that contains equal amounts of HA and A^{-} (0.90 mmol), and so the pH equals pK_{a2} of the methionine cation H_{2}A^{+}.
pH = pK_{a2} =-log (6.2\times 10^{-10})=9.21(c) The amount of NaOH added to the 1.80 mmol of H_{2}A^{+} present initially is (35.0 mL) × (0.0900 mmol/mL) = 3.15 mmol. The added base is enough to convert all of the H_{2}A^{+} to 1.80 mmol of HA and then convert 3.15 – 1.80 = 1.35 mmol of the resulting HA to 1.35 mmol of A^{-}.
Table 3
After neutralization, we have an HA–A^{-} buffer solution that contains 0.45 mmol of HA and 1.35 mmol of A^{-}. We can use the Henderson–Hasselbalch equation to calculate the pH:
pH = pK_{a2} + log\frac{[Base]}{[Acid]}=9.21 + log\frac{1.35}{0.45}=9.69Table 1
| Neutralization Reaction | H_{2}A^{+}(aq) | + | OH^{-}(aq) | \overset{100\%}{\longrightarrow} | H_{2}O(l) | + | HA(aq) |
| Before reaction (mol) | 1.80 | 1.80 | 0 | ||||
| Change (mol) | -1.80 | -1.80 | + 1.80 | ||||
| After reaction (mol) | ∼ 0 | ∼ 0 | 1.80 |
Table 2
| Neutralization Reaction | HA(aq) | + | OH^{-}(aq) | \overset{100\%}{\longrightarrow} | H_{2}O(l) | + | A^{-}(aq) |
| Before reaction (mol) | 1.80 | 0.90 | 0 | ||||
| Change (mol) | -0.90 | -0.90 | + 0.90 | ||||
| After reaction (mol) | 0.90 | ∼ 0 | 0.90 |
Table 3
| Neutralization Reaction | HA(aq) | + | OH^{-}(aq) | \overset{100\%}{\longrightarrow} | H_{2}O(l) | + | A^{-}(aq) |
| Before reaction (mol) | 1.80 | 1.35 | 0 | ||||
| Change (mol) | -1.35 | -1.35 | + 1.35 | ||||
| After reaction (mol) | 0.45 | ∼ 0 | 1.35 |