JEE Chemistry 2025 Question with Solution

Given: A solution contains $1 \, \text{M}$ $A^{2+}$ and $1 \, \text{M}$ $B^{3+}$ ions. Also, $K_{sp}[A(OH)_2] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_3] = 27 \times 10^{-18}$.

Find: Which hydroxide precipitates first when $NH_4OH$ is added gradually.

Precipitation starts for the ion that requires the lower hydroxide ion concentration to satisfy its solubility product.

For $A(OH)_2$:

$$A(OH)_2 \rightleftharpoons A^{2+} + 2OH^-$$

Its solubility product expression is

$$K_{sp}[A(OH)_2] = [A^{2+}][OH^-]^2$$

Using $[A^{2+}] = 1 \, \text{M}$,

$$9 \times 10^{-10} = 1 \times [OH^-]^2$$ $$[OH^-] = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \, \text{M}$$

For $B(OH)_3$:

$$B(OH)_3 \rightleftharpoons B^{3+} + 3OH^-$$

Its solubility product expression is

$$K_{sp}[B(OH)_3] = [B^{3+}][OH^-]^3$$

Using $[B^{3+}] = 1 \, \text{M}$,

$$27 \times 10^{-18} = 1 \times [OH^-]^3$$ $$[OH^-] = \sqrt[3]{27 \times 10^{-18}} = 3 \times 10^{-6} \, \text{M}$$

Now compare the hydroxide ion concentrations required for precipitation:

$$3 \times 10^{-6} \, \text{M} < 3 \times 10^{-5} \, \text{M}$$

So $B(OH)_3$ reaches its precipitation condition first.

Therefore, $B(OH)_3$ precipitates before $A(OH)_2$. The correct option is A.

Given: $NH_4OH$ is added gradually to a solution containing $1 \, \text{M}$ $A^{2+}$ and $1 \, \text{M}$ $B^{3+}$.

Find: Which hydroxide starts precipitating first.

The hydroxide that needs the smaller critical value of $[OH^-]$ will precipitate earlier.

For $A(OH)_2$, precipitation begins when

$$[A^{2+}][OH^-]^2 = K_{sp}[A(OH)_2]$$

Substituting the given values,

$$1 \times [OH^-]^2 = 9 \times 10^{-10}$$ $$[OH^-]^2 = 9 \times 10^{-10}$$ $$[OH^-] = 3 \times 10^{-5} \, \text{M}$$

For $B(OH)_3$, precipitation begins when

$$[B^{3+}][OH^-]^3 = K_{sp}[B(OH)_3]$$

Substituting the given values,

$$1 \times [OH^-]^3 = 27 \times 10^{-18}$$ $$[OH^-]^3 = 27 \times 10^{-18}$$ $$[OH^-] = 3 \times 10^{-6} \, \text{M}$$

Since the required hydroxide ion concentration is smaller for $B(OH)_3$, it precipitates first on gradual addition of $NH_4OH$.

Therefore, the correct option is A.