Ionic Equilibrium JEE 2023: Solubility and pKa Solution

What is the JEE Advanced 2023 Paper 1 ionic equilibrium solubility question?

Option B) 4 is correct. In this JEE Advanced 2023 ionic equilibrium question, total salt solubility differs from free anion concentration because some anion forms HX. This is a Chemistry, Ionic Equilibrium, JEE Advanced 2023 Paper 1 single-correct MCQ.

A sparingly soluble salt, MX, contains the conjugate base of the weak acid HX. Lowering the solution’s pH from 7 to 2 raises its solubility as follows:

S_{p H 7} = 10^{- 4} m o l L^{- 1}, S_{p H 2} = 10^{- 3} m o l L^{- 1} .

Find the pKa of HX. Choose one answer:

The question bank rates this medium, with an expected solving time of 120 seconds. That is a practice estimate, not an official exam allowance.

How do you obtain Ksp from the solubility at pH 7?

The official method neglects protonation of the anion at pH 7, making both free-ion concentrations approximately equal to the salt’s solubility. This approximation needs checking after finding pKa; it does not apply at every pH.

The dissolution equilibrium and initial solubility are:

M X (s) ⇌ M^{+} + X^{-}

S = 10^{- 4} m o l L^{- 1} .

Each dissolved formula unit supplies one cation and one anion. If negligible anion becomes HX, then:

[M^{+}] \approx [X^{-}] \approx S .

Therefore:

K_{s p} = [M^{+}] [X^{-}] \approx S^{2} = (10^{- 4})^{2} = 10^{- 8} .

Lowering pH consumes free anion by forming HX, allowing more salt to dissolve. At the same temperature, Ksp stays fixed: solubility and individual ion concentrations can change, but their equilibrium product cannot.

Why is free anion concentration different from solubility at pH 2?

Total solubility counts all dissolved salt, but free anion concentration excludes the anion converted to HX. At pH 2, obtain the free anion from Ksp first, then use the material balance to find HX.

Define the new solubility:

S^{'} = 10^{- 3} m o l L^{- 1} .

The 1:1 salt stoichiometry gives:

[M^{+}] = S^{'} = 10^{- 3} m o l L^{- 1} .

Using the approximate Ksp obtained above:

10^{- 8} \approx (10^{- 3}) [X^{-}]

[X^{-}] \approx 10^{- 5} m o l L^{- 1} .

Every dissolved anion unit must be either free anion or HX. The material balance therefore gives:

S^{'} = [X^{-}] + [H X]

[H X] \approx 10^{- 3} - 10^{- 5} = 9.9 \times 10^{- 4} m o l L^{- 1} .

The distinction is:

\underset{total solubility}{\underset{⏟}{S^{'} = 10^{- 3} m o l L^{- 1}}} \underset{free anion}{\underset{⏟}{[X^{-}] \approx 10^{- 5} m o l L^{- 1}}} .

Relative to pH 7, the cation concentration increases tenfold while free anion concentration decreases approximately tenfold. Their product stays unchanged, which checks the calculation.

How does the calculation give pKa 4, and is the pH 7 approximation valid?

The pH 2 concentrations give pKa approximately 4, so option B) 4 follows. This also validates the starting approximation: at pH 7, protonated anion is only about 0.1% of free anion.

For the weak acid: $H X ⇌ H^{+} + X^{-}$

K_{a} = \frac{[H^{+}] [X^{-}]}{[H X]} .

At pH 2:

[H^{+}] = 10^{- 2} m o l L^{- 1} .

Substitution gives:

K_{a} \approx \frac{(10^{- 2}) (10^{- 5})}{9.9 \times 10^{- 4}} \approx 1.01 \times 10^{- 4} \approx 1.0 \times 10^{- 4} .

Thus:

p K_{a} = - \log_{10} K_{a} \approx 4 .

The small difference from exactly 4 is consistent with the approximation used, not evidence for another option. Check the initial solution at pH 7:

\frac{[H X]}{[X^{-}]} = \frac{[H^{+}]}{K_{a}} \approx \frac{10^{- 7}}{10^{- 4}} = 10^{- 3} .

The protonated amount is about 0.1% of the free anion. Neglecting it when calculating Ksp from the initial solubility is justified.

How does incorrectly holding free anion concentration constant produce option A?

Holding free anion concentration at its pH 7 value gives pKa near 3, but violates Ksp. The incorrect assumption is that acidification leaves free anion unchanged even though more salt dissolves:

[X^{-}]_{w r o n g} = 10^{- 4} m o l L^{- 1} .

The resulting material balance gives:

[H X]_{w r o n g} = 10^{- 3} - 10^{- 4} = 9 \times 10^{- 4} m o l L^{- 1} .

That leads to:

K_{a, w r o n g} = \frac{(10^{- 2}) (10^{- 4})}{9 \times 10^{- 4}} \approx 1.11 \times 10^{- 3}

p K_{a, w r o n g} \approx 2.95 .

This points to option A) 3. But the assumed ion concentrations contradict the solubility product:

[M^{+}] [X^{-}] = (10^{- 3}) (10^{- 4}) = 10^{- 7} \neq K_{s p} \approx 10^{- 8} .

Ksp remains fixed, not either ion concentration separately. Repair the method by using the new cation concentration and the fixed Ksp to calculate free anion concentration before applying the material balance.

What two related questions can you solve using the same method?

Calculate the anion percentages at pH 2 and the solubility at a maintained pH of 4. Both questions below are original practice based on this PYQ, not additional verified past-year questions.

At pH 2, what percentage of dissolved anion is free and what percentage is HX?

The dissolved anion is approximately 1% free anion and 99% HX. Divide each form’s concentration by total solubility, not by the concentration of the other form:

% free X^{-} \approx \frac{10^{- 5}}{10^{- 3}} \times 100 = 1 % .

% H X \approx 100 % - 1 % = 99 % .

These percentages describe dissolved anion units. They do not describe the fraction of the original solid that dissolved.

What is the solubility at a maintained pH of 4?

Using the calculated Ksp and pKa under the same assumptions, the solubility is approximately:

s \approx 1.41 \times 10^{- 4} m o l L^{- 1} .

The constants used are:

K_{s p} \approx 10^{- 8}, p K_{a} \approx 4 .

At pH 4:

[H^{+}] = 10^{- 4} m o l L^{- 1} \approx K_{a} .

The acid equilibrium therefore gives approximately equal concentrations of the two anion forms. Within this approximation, the material balance and salt stoichiometry give:

[H X] \approx [X^{-}], [M^{+}] = s, [X^{-}] \approx \frac{s}{2} .

Hence:

K_{s p} \approx s (\frac{s}{2}) = \frac{s^{2}}{2} \approx 10^{- 8}

s \approx \sqrt{2 \times 10^{- 8}} \approx 1.41 \times 10^{- 4} m o l L^{- 1} .

This lies between the given pH 7 and pH 2 solubilities: protonation is stronger than at pH 7 but weaker than at pH 2. Before accepting your result, check both the anion material balance and the free-ion product.

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Frequently asked questions

What is the answer to the JEE Advanced 2023 Paper 1 ionic equilibrium solubility question?

The correct answer is option B) 4. At pH 2, the free anion concentration is approximately 10^-5 mol/L and the HX concentration is 9.9 × 10^-4 mol/L. Substituting these into Ka = [H+][X-]/[HX] gives Ka approximately 1.01 × 10^-4, so pKa is approximately 4.

Why does lowering pH increase the solubility of MX?

The anion X- is the conjugate base of the weak acid HX, so added acid converts some free X- into HX. This allows more MX to dissolve while maintaining the same Ksp at a fixed temperature. Total solubility therefore increases even though the free anion concentration decreases.

Why can we use Ksp = S squared at pH 7 in this question?

At pH 7, protonation of X- is negligible, so both free-ion concentrations are approximately equal to the solubility S. The calculated pKa of about 4 gives [HX]/[X-] approximately 10^-3 at pH 7. Thus, HX is only about 0.1% of the free anion, validating Ksp approximately equal to S squared.

Why is pKa 3 wrong in the JEE 2023 solubility question?

A pKa near 3 results from incorrectly keeping [X-] fixed at its pH 7 value of 10^-4 mol/L. At pH 2, [M+] is 10^-3 mol/L, so that assumption gives an ion product of 10^-7 instead of Ksp approximately 10^-8. Keep Ksp fixed and calculate the new free anion concentration before applying the anion material balance.