Question 16.6: What [NH3]/[NH4^+] ratio is required for a buffer solution t......
What [NH_{3}]/[NH_{4} ^{+}] ratio is required for a buffer solution that has pH = 7.00? Is a mixture of NH_{3} and NH_{4}Cl a good choice for a buffer having pH = 7.00?
STRATEGY
Find the pK_{a} value for NH_{4} ^{+} from the tabulated K_{b} value for NH_{3} (Appendix C). Use the Henderson–Hasselbalch equation to calculate the [NH_{3}]/[NH_{4} ^{+}] ratio for the desired pH.
| IDENTIFY | |
| Known | Unknown |
| pH = 7.00 | Ratio of NH_{3}/NH_{4} ^{+} |
| Identity of weak acid and conjugate base (NH_{4}Cl-NH_{3}) | |
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}=-logK_{a}=-log (5.6\times 10^{-10})=9.25
Rearrange the Henderson–Hasselbalch equation to obtain an expression for the relative amounts of NH_{3} and NH_{4} ^{+} in a solution having pH = 7.00:
log\frac{[Base]}{[Acid]}=pH – pK_{a} =7.00-9.25=-2.25Therefore,
\frac{[NH_{3}]}{[NH_{4} ^{+}]}=antilog (-2.25)=10^{-2.25}=5.6\times 10^{-3}For a typical value of [NH_{4} ^{+}]—say, 1.0 M—the NH_{3} concentration would have to be very small (0.0056 M). Such a solution is a poor buffer because it has little capacity to absorb added acid. Also, because the [NH_{3}]/[NH_{4} ^{+}] ratio is far from 1.0, addition of a small amount of H_{3}O^{+} or OH^{-} will result in a large change in the pH.