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Question #44118

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The Question

A concentration cell at 298 K is constructed with two standard hydrogen electrodes ( $P_{\ce{H2}} = 1 \text{ bar}$ ):

$\ce{Pt | H2(g) | Solution A || Solution B | H2(g) | Pt}$

Solution A consists of 1.0 L of a $0.10 \text{ M}$ aqueous solution of $\ce{NaIn}$ , the sodium salt of a weak acid indicator $\ce{HIn}$ ( $K_a = 2.0 \times 10^{-2}$ ). Solution B consists of 1.0 L of a $0.12 \text{ M}$ aqueous solution of $\ce{HCl}$ . The electrodes are connected by a wire, and the cell is allowed to discharge completely until the cell potential drops to zero.

Assuming no volume changes and ideal behavior, what is the final concentration of $\ce{H+}$ in Solution B, and the percentage of the indicator that remains in the ionized form ( $\ce{In^-}$ ) in Solution A at equilibrium?

Options

A.
$[\ce{H+}] = 0.03 \text{ M}$ , $40\%$
B.
$[\ce{H+}] = 0.05 \text{ M}$ , $30\%$
C.
$[\ce{H+}] = 0.01 \text{ M}$ , $0\%$
D.
$[\ce{H+}] = 0.03 \text{ M}$ , $60\%$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Titration Indicators|Degree of Dissociation (Alpha) approximation|Concentration Cells

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Nernstian electrochemical equilibrium and exact ionic mass balance.

The 'Hidden Variable' is the free $\ce{H+}$ concentration in the indicator half-cell.

Students typically assume that all transferred $\ce{H+}$ is consumed by the weak base ( $\ce{In-}$ ), using the approximation $[\ce{HIn}] \approx x$ .

However, because $K_a$ is relatively large ( $2.0 \times 10^{-2}$ ), a significant portion of the transferred $\ce{H+}$ remains free to satisfy the Nernstian equilibrium ( $E_{cell} = 0 \implies [\ce{H+}]_A = [\ce{H+}]_B$ ).

Failing to account for this free $\ce{H+}$ in the mass balance leads to an incorrect quadratic formulation (Option B).

Treating the discharge as a simple 'limiting reagent' complete reaction ignores the reversible equilibrium state entirely (Option C).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Titration Indicators
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free