Question

In a calibrated pH meter comprising of glass electrode-standard calomel electrode, the potential for a buffer solution of pH 4.01 is measured as 0.814 V at 25\(^{\circ}\)C. For a \(4.0 \times 10^{-3}\) M solution of acetic acid, the measured potential (in V) is (rounded off to three decimal places).
(Given: \(K_a\) of acetic acid at 25\(^{\circ}\)C = \(1.75 \times 10^{-5}\); \(2.303RT/F = 0.059\); Assume: \(a_{H^+} = [H^+]\))

Hint:

First find the pH of the \(4.0\times10^{-3}\) M acetic acid solution by solving the \(K_a\) quadratic exactly (not the simple square-root approximation), then use the Nernstian line \(E=K-0.059\,pH\) calibrated at pH 4.01.

Answer 1

Correct Answer: 0.839

Step 1: Find \([H^+]\) for the acetic acid solution from \(K_a\).
Acetic acid is a weak acid, \(\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}\), with \(K_a = \dfrac{[H^+][CH_3COO^-]}{[CH_3COOH]}\). Let \(x=[H^+]\) at equilibrium, starting from \(C=4.0\times10^{-3}\) M:
\[ K_a = \frac{x^2}{C-x} \] \[ x^2 + K_a x - K_a C = 0 \]
Step 2: Solve the quadratic (the simple \(x=\sqrt{K_aC}\) approximation is not accurate enough here since \(x/C\) works out to about 6%).
\[ x^2 + (1.75\times10^{-5})x - (1.75\times10^{-5})(4.0\times10^{-3}) = 0 \] \[ x^2 + 1.75\times10^{-5}x - 7.00\times10^{-8} = 0 \] \[ x = \frac{-1.75\times10^{-5}+\sqrt{(1.75\times10^{-5})^2+4(7.00\times10^{-8})}}{2} \] \[ x = \frac{-1.75\times10^{-5}+\sqrt{2.803\times10^{-7}}}{2} = \frac{-1.75\times10^{-5}+5.2944\times10^{-4}}{2} \] \[ x = [H^+] = 2.5597\times10^{-4}\ \mathrm{M} \]
Step 3: Convert to pH.
\[ pH = -\log_{10}(2.5597\times10^{-4}) = 4 - \log_{10}(2.5597) = 4 - 0.4082 = 3.5918 \]
Step 4: Use the calibration line of the glass-calomel cell.
A glass-calomel pH cell responds Nernstianly to pH: \(E = K - (2.303RT/F)\,pH\), so \(E\) falls by 0.059 V for every unit rise in pH. Use the pH 4.01 calibration point to find the cell constant \(K\):
\[ 0.814 = K - (0.059)(4.01) \] \[ K = 0.814 + 0.23659 = 1.05059\ \mathrm{V} \]
Step 5: Apply the same line to the acetic acid solution's pH.
\[ E = K - 0.059\,(pH) = 1.05059 - 0.059(3.5918) \] \[ E = 1.05059 - 0.21192 = 0.83867\ \mathrm{V} \]
Final Answer:
Rounded to three decimal places, the measured potential is \[ \boxed{E \approx 0.839\ \mathrm{V}} \]