Question

Solubility of a $M_2S$ salt is $3.5 \times 10^{–6}$ , then find out solubility product.

• A. $1.7 \times10^{–6}$
• B. $1.7 \times 10^{–16}$
• C. $1.7\times 10^{–18}$
• D. $1.7 \times 10^{–12}$

Answer 1

Answer: (B)

$1.7 \times 10^{–16}$

Answer 2

To find the solubility product ($Ksp$) of a sparingly soluble salt like $M_2S$ (where M represents a metal cation and S represents a sulfide anion), you need to know the solubility (S) of the salt.

Given that the solubility of $M_2S$ is $3.5 \times 10^{–6}$, you can set up an equilibrium expression for the dissolution of the salt:

$M_2S (s) ⇌ 2M^+ (aq) + S^{²-} (aq)$

Now, you can write the expression for the solubility product ($Ksp$) Using the concentrations of the ions involved:$Ksp$ = $[M^+]^2[S^{2-}]$
Since $M_2S$ dissociates to form $2 M^+$ ions and $1 S^{2-}$ ion: $Ksp$ = $(2S)^{2}(S) = 4S^{3}$

Now, substitute the solubility value ($S = 3.5 \times 10^{–6}$) into the equation for$Ksp$: $Ksp$ = $4(3.5 \times 10^{–6})^3 = 1.7 \times 10^{^-16}$
So, the solubility product ($Ksp$) of $M_2S$ is indeed $1.7 \times 10^{–16}$.

So, The correct option is (B): $1.7 \times 10^{–16}.$