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

Since the pH meter is Nernstian, we do not need to find the full calibration constant $K$ at all. A change in pH just produces a proportional drop in the measured potential, so the second reading can be found directly from the first one and the pH difference.

1. Get the acetic acid pH first: solve the acid dissociation quadratic $x^2 + K_a x - K_aC = 0$ with $K_a=1.75\times10^{-5}$, $C=4.0\times10^{-3}$ M (the simple $\sqrt{K_aC}$ shortcut is not accurate enough, since $x$ turns out to be about 6% of $C$). This gives $[H^+]=2.560\times10^{-4}$ M, so $pH = -\log(2.560\times10^{-4}) = 3.592$.
2. Use the Nernst response directly: for a glass electrode combination cell, $E = K - 0.059\,pH$, so between any two readings $E_2 - E_1 = -0.059(pH_2-pH_1)$. There is no need to solve for $K$ separately.
3. Plug in the two states: state 1 is the pH 4.01 buffer at $E_1=0.814$ V, state 2 is the acid solution at $pH_2=3.592$. The pH change is $pH_2-pH_1 = 3.592-4.01 = -0.418$.
4. Apply the shift: $E_2 = 0.814 - 0.059(-0.418) = 0.814 + 0.02466 = 0.83866$ V.

Let's summarize:

• The calibration point only fixes the line's slope-intercept relationship; you can jump straight between any two points on it using $\Delta E = -0.059\,\Delta(pH)$.
• A lower pH (more acidic than the buffer) gives a higher potential here, since $E$ and pH move in opposite directions on this electrode.

So the measured potential, rounded to three decimal places, is $\boxed{0.839\ \text{V}}$.