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Question #44115

Difficulty: 8/10 | 🟢 Free

The Question

The effective Van't Hoff factor ( $i$ ) for a $0.05 \text{ M}$ aqueous solution of a weak diprotic acid $\ce{H2A}$ is $1.80$ at $298 \text{ K}$ . If the second ionization constant of the acid is $K_{a2} = 2.0 \times 10^{-2}$ , calculate the first ionization constant ( $K_{a1}$ ) of the acid. Assume ideal behavior of all dissolved particles and neglect the autoionization of water.

Options

A.
$4.0 \times 10^{-2}$
B.
$8.0 \times 10^{-2}$
C.
$1.6 \times 10^{-1}$
D.
$2.0 \times 10^{-2}$

Check Your Answer

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

Successive Equilibrium Constants|Van’t Hoff Factor (i) Application

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Hidden Variable' of total particle count in successive equilibria.

Students typically fall into two traps: (1) Treating the acid as monoprotic (ignoring $K_{a2}$ ), which leads to $[\ce{H+}] = [\ce{HA-}] = 0.04 \text{ M}$ and an incorrect $K_{a1}$ of $1.6 \times 10^{-1}$ . (2) Misapplying the mass balance by assuming $[\ce{H2A}] = C - [\ce{H+}]$ , a formula only valid for monoprotic acids, leading to $8.0 \times 10^{-2}$ .

The key to unlocking the problem is recognizing that for any pure polyprotic acid, the total particle concentration simplifies elegantly to $P = C + [\ce{H+}]$ , bypassing the need to solve intractable cubic equations.

Quick Facts About This Question

Question Information

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