Question 16.5: Use the Henderson–Hasselbalch equation to calculate the pH o......
Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution that is 0.45 M in NH_{4}Cl and 0.15 M in NH_{3}.
STRATEGY
We’ve already solved this problem by another method in Worked Example 16.2. Now that we’ve discussed equilibria in buffer solutions, though, we can use the Henderson-Hasselbalch equation as a shortcut. Since NH_{4} ^{+} is the weak acid in the NH_{4} ^{+}–NH_{3} buffer solution, we need to find the pK_{a} for NH_{4} ^{+} from the tabulated K_{b} value for NH_{3} (Appendix C). Then, substitute [NH_{3}], [NH_{4} ^{+}], and the pK_{a} value into the Henderson–Hasselbalch equation to find the pH.
| IDENTIFY | |
| Known | Unknown |
| Concentration of weak acid and conjugate base (0.45 M NH_{4}Cl and 0.15 M NH_{3}) |
pH |
Since K_{b} for NH_{3} is 1.8\times 10^{-5}, K_{a} for NH_{4} ^{+} is 5.6\times 10^{-10} and pK_{a} = 9.25.
K_{a}=\frac{K_{w}}{K_{b}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.6\times 10^{-10}pK_{a}=-log K_{a}=-log (5.6\times 10^{-10})=9.25
Since [base] = [NH_{3}] = 0.15 M and [acid] = [NH_{4} ^{+}] = 0.45 M,
pH = pK_{a} + log\frac{[Base]}{[Acid]}=9.25 + log(\frac{0.15}{0.45})=9.25-0.48=8.77The pH of the buffer solution is 8.77.