Problem
A buffer is prepared from acetic acid \(\mathrm{HA}\) and sodium acetate \(\mathrm{A^-}\) so that the equilibrium mixture has \([\mathrm{HA}]=0.150\,\mathrm{M}\) and \([\mathrm{A^-}]=0.100\,\mathrm{M}\). Given \(K_a(\text{acetic acid})=1.8\times 10^{-5}\), determine the pH using the henderson hasselbalch equation.
Acid–base pair: \(\mathrm{HA}\) (weak acid) and \(\mathrm{A^-}\) (conjugate base).
Key idea: buffer pH depends mainly on \(pK_a\) and the ratio \([\mathrm{A^-}]/[\mathrm{HA}]\).
Solution
1) Start from the weak-acid equilibrium expression
For \(\mathrm{HA(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{A^-(aq)}\), the acid dissociation constant is
Rearrange to isolate \([\mathrm{H^+}]\):
2) Convert to logarithmic form (Henderson–Hasselbalch)
Take \(-\log_{10}\) of both sides and use \(\mathrm{pH}=-\log_{10}([\mathrm{H^+}])\) and \(pK_a=-\log_{10}(K_a)\):
Practical use: when both \(\mathrm{HA}\) and \(\mathrm{A^-}\) are present in appreciable amounts, the ratio \([\mathrm{A^-}]/[\mathrm{HA}]\) sets the pH relative to \(pK_a\). If \([\mathrm{A^-}]=[\mathrm{HA}]\), then \(\mathrm{pH}=pK_a\).
3) Compute \(pK_a\) from \(K_a\)
4) Compute the concentration ratio and the pH
Evaluate the ratio and its base-10 logarithm:
Substitute into the Henderson–Hasselbalch equation:
| Quantity | Expression | Value |
|---|---|---|
| \(pK_a\) | \(-\log_{10}(K_a)\) | \(4.7447\) |
| Ratio | \([\mathrm{A^-}]/[\mathrm{HA}]\) | \(0.6667\) |
| Log ratio | \(\log_{10}([\mathrm{A^-}]/[\mathrm{HA}])\) | \(-0.1761\) |
| pH | \(pK_a+\log_{10}([\mathrm{A^-}]/[\mathrm{HA}])\) | \(4.57\) |
5) Interpretation check
- Since \([\mathrm{A^-}]/[\mathrm{HA}]<1\), the logarithm is negative, so \(\mathrm{pH}
- The computed pH \(4.57\) is slightly below \(4.74\), consistent with having more \(\mathrm{HA}\) than \(\mathrm{A^-}\).
Visualization
Final answer
With \(K_a=1.8\times 10^{-5}\) (\(pK_a=4.7447\)) and \([\mathrm{A^-}]/[\mathrm{HA}]=0.6667\), the henderson hasselbalch equation gives \(\mathrm{pH}\approx 4.57\).