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Question #44168

Difficulty: 8/10 | 🟢 Free

The Question

Consider a highly concentrated $2.5 \text{ M}$ aqueous solution of a sodium carboxylate salt, $\ce{NaA}$ , at $298 \text{ K}$ . The thermodynamic acid dissociation constant of the weak acid $\ce{HA}$ is $K_a^\circ = 2 \times 10^{-6}$ . Due to intense hydrogen-bonding competition and the formation of primary solvation shells around the high concentration of $\ce{Na^+}$ and $\ce{A^-}$ ions, the activity of 'free' water is reduced to $a_w = 0.8$ . Assuming the activity coefficients of all ionic and neutral solute species are approximately unity ( $\gamma \approx 1$ ) and standard $K_w^\circ = 10^{-14}$ , calculate the true degree of hydrolysis ( $\alpha$ ) of the salt and the thermodynamic activity of hydrogen ions ( $a_{\ce{H^+}}$ ) in this solution.

Options

A.
$\alpha = 4.0 \times 10^{-5}, \ a_{\ce{H^+}} = 8.0 \times 10^{-11}$
B.
$\alpha = 4.47 \times 10^{-5}, \ a_{\ce{H^+}} = 8.94 \times 10^{-11}$
C.
$\alpha = 4.0 \times 10^{-5}, \ a_{\ce{H^+}} = 1.0 \times 10^{-10}$
D.
$\alpha = 5.0 \times 10^{-5}, \ a_{\ce{H^+}} = 8.0 \times 10^{-11}$

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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

Hydrogen Bonding Competition|Salt Hydrolysis Logic|Degree of Dissociation (α) approximation

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Hidden Variable' of water activity ( $a_w$ ) in highly concentrated solutions.

Students typically memorize $\alpha = \sqrt{K_h/C}$ and $K_h = K_w/K_a$ , assuming $a_w = 1$ (ideal dilute solution).

In concentrated solutions, extensive H-bonding and solvation by ions reduce 'free' water activity.

This requires modifying both the apparent hydrolysis constant ( $K_h' = K_h^\circ \cdot a_w$ ) and the apparent ionic product of water ( $K_w' = K_w^\circ \cdot a_w$ ).

Failing to account for $a_w$ leads to Option B.

Applying it to $\alpha$ but forgetting it for $K_w$ leads to Option C.

Incorrectly placing $a_w$ in the denominator of $K_h$ leads to Option D.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Hydrogen Bonding Competition
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free