At 30°C the solubility of PbI$_2$ salt in 0.2 M KI solution will be X, if the solubility product of PbI$_2$ at 30°C is $2.4 \times 10^{-8}$. Identify the value of X.

The dissociation of $PbI_2$ is given by:

$PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$

Let the solubility of $PbI_2$ in $0.2$ M $KI$ solution be $s$.

The concentration of $I^-$ ions from $KI$ is $0.2$ M. Since $KI$ is a strong electrolyte, it completely dissociates. The total concentration of $I^-$ ions is $(2s + 0.2)$ M. Because $K_{sp}$ is very small, $2s$ can be neglected compared to $0.2$.

$[I^-] \approx 0.2$ M

The expression for the solubility product is:

$K_{sp} = [Pb^{2+}][I^-]^2$

Substituting the given values:

$2.4 \times 10^{-8} = s \times (0.2)^2$

$2.4 \times 10^{-8} = s \times 0.04$

$s = \dfrac{2.4 \times 10^{-8}}{0.04} = 6.0 \times 10^{-7}$ M

Thus, the solubility $X$ is $6.0 \times 10^{-7}$ M.

Answer: $6.0 \times 10^{-7}$M