JEE Chemistry 2025 Question with Solution

Given: $K_{sp}$ for $Cr(OH)_3$ is $1.6 \times 10^{-30}$.

Find: The molar solubility $s$ of $Cr(OH)_3$ in water.

The dissolution equilibrium is:

$$Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3OH^- (aq)$$

If the molar solubility is $s$, then

• $[Cr^{3+}] = s$
• $[OH^-] = 3s$

Using the solubility product expression:

$$K_{sp} = [Cr^{3+}][OH^-]^3$$

Substitute the concentrations:

$$K_{sp} = (s)(3s)^3 = 27s^4$$

Now use the given value:

$$1.6 \times 10^{-30} = 27s^4$$

So,

$$s^4 = \frac{1.6 \times 10^{-30}}{27}$$

Taking the fourth root,

$$s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$$

Therefore, the molar solubility is $s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$ and the correct option is A.

Given: $Cr(OH)_3$ dissolves with $K_{sp} = 1.6 \times 10^{-30}$.

Find: The expression for molar solubility $s$.

For each mole of $Cr(OH)_3$ that dissolves, it forms one mole of $Cr^{3+}$ and three moles of $OH^-$.

Hence, if solubility is $s$:

$$[Cr^{3+}] = s, \qquad [OH^-] = 3s$$

Now write the equilibrium expression:

$$K_{sp} = [Cr^{3+}][OH^-]^3$$

Substituting,

$$K_{sp} = s(3s)^3$$

Evaluate the cube term:

$$(3s)^3 = 27s^3$$

Therefore,

$$K_{sp} = s \cdot 27s^3 = 27s^4$$

With $K_{sp} = 1.6 \times 10^{-30}$,

$$27s^4 = 1.6 \times 10^{-30}$$

Rearranging,

$$s^4 = \frac{1.6 \times 10^{-30}}{27}$$

Thus,

$$s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$$

So the correct option is A.