The molar solubility of Zn(OH)$$_2$$ in 0.1M NaOH solution is $$x \times 10^{-18}$$ M. The value of x is _________. (Nearest integer)
Given: The solubility product of Zn(OH)$$_2$$ is $$2 \times 10^{-20}$$

The dissolution equilibrium for zinc hydroxide is
$$\text{Zn(OH)}_2(s) \rightleftharpoons \text{Zn}^{2+}(aq) + 2\;\text{OH}^-(aq)$$

For every mole (solubility $$s$$) of $$\text{Zn(OH)}_2$$ that dissolves:
  • $$[\text{Zn}^{2+}] = s$$
  • $$[\text{OH}^-] = 0.10\;\text{M (from NaOH)} + 2s$$

The given solubility product expression is
$$K_{sp} = [\text{Zn}^{2+}]\,[\text{OH}^-]^2$$

Substituting the equilibrium concentrations:
$$2 \times 10^{-20} = s\,(0.10 + 2s)^2 \; -(1)$$

The added hydroxide $$2s$$ is extremely small compared with the fixed $$0.10\;\text{M}$$ (because solubilities of such sparingly soluble hydroxides are usually of the order of $$10^{-17}\;\text{M}$$ or less). Hence,
$$0.10 + 2s \approx 0.10$$

Applying this approximation to $$(1)$$:
$$2 \times 10^{-20} \approx s\,(0.10)^2$$
$$2 \times 10^{-20} = s\,(1 \times 10^{-2})$$

Solving for $$s$$:
$$s = \frac{2 \times 10^{-20}}{1 \times 10^{-2}} = 2 \times 10^{-18}\;\text{M}$$

Thus the molar solubility can be written as $$x \times 10^{-18}\;\text{M}$$ with $$x = 2$$.

Answer (nearest integer): $$2$$.