Question

17.74 Mercury(II) ion is often precipitated as mercury(II) sulfide in qualitative analysis. Using the solubility product con- stant (Table 17.1), calculate the molar solubility of mercury(II) Sulfide in water, assuming that no other reactions occur.

Answer 1

HgS in water dissociates as

HgS (s)  $\small \overset{H_{2}O}{\rightleftharpoons }$ Hg²⁺ (aq) + S^-2 (aq)

let solubility of HgS = S ( mol/ L or M)

then, [Hg²⁺] = [S^2-] = S ( mol/ L or M)

hence , K = $\small \frac{[Hg^{2+}][S^{2-}]}{{[HgS]}}$

as HgS is sparingly soluble salt so,

or, K [HgS] = constant = K_sp

Ksp is called solubility product.

hence, Ksp = [Hg²⁺] [S^2-] = S $\small \times$ S

or, K_sp = S²

theoretical value of Ksp of HgS = 1 $\small \times$10^-54

then

1 $\small \times$10^-54 = S²

or, S = (1 $\small \times$10^-54)^0.5 = 1 $\small \times$10^-27 M

hence molar solubility of mercury(II) sulfide in water =  1 $\small \times$10^-27 M