Question

At 25$^\circ$C, the solubility product of MCl is $1 \times 10^{-10}$. What is its molar solubility in 0.1 M NaCl solution at same temperature?

• A. 0.1 M
• B. 0.05 M
• C. $10^{-9}$ M
• D. $10^{-10}$ M

Hint:

Write the dissolution equilibrium and $K_{sp}$ expression for the sparingly soluble salt MCl.
Identify the common ion (Cl$^-$) and its initial concentration from the strong electrolyte (NaCl).
Let $s$ be the molar solubility of MCl. Set up an ICE table or directly write equilibrium concentrations: $[\text{M}^+]=s$, $[\text{Cl}^-]=[\text{Cl}^-]_{initial} + s$.
Substitute into $K_{sp}$ expression: $K_{sp} = s([\text{Cl}^-]_{initial} + s)$.
If $K_{sp}$ is small and $[\text{Cl}^-]_{initial}$ is significant, $s$ will be small. Use the approximation $[\text{Cl}^-]_{initial} + s \approx [\text{Cl}^-]_{initial}$.
Solve for $s$ and verify the approximation.

Answer 1

The Correct Option is C

Let MCl be a sparingly soluble salt. Its dissolution equilibrium in water is: MCl(s) $\rightleftharpoons$ M$^+$(aq) + Cl$^-$(aq) The solubility product constant ($K_{sp}$) is given by: $K_{sp} = [\text{M}^+][\text{Cl}^-]$ Given $K_{sp} = 1 \times 10^{-10}$ at $25^\circ\text{C}$. We need to find the molar solubility ($s$) of MCl in a $0.1 \text{ M NaCl}$ solution. NaCl is a strong electrolyte and dissociates completely: NaCl(aq) $\rightarrow$ Na$^+$(aq) + Cl$^-$(aq). So, a $0.1 \text{ M NaCl}$ solution provides an initial concentration of Cl$^-$ ions: $[\text{Cl}^-]_{initial} = 0.1 \text{ M}$. This is an example of the common ion effect. Let $s$ be the molar solubility of MCl in the NaCl solution. This means that $s$ moles of MCl dissolve per liter of solution. When $s$ moles/L of MCl dissolve, it produces $s$ moles/L of M$^+$ ions and $s$ moles/L of Cl$^-$ ions from MCl. So, $[\text{M}^+] = s$. The total concentration of Cl$^-$ ions in the solution will be the sum of Cl$^-$ from NaCl and Cl$^-$ from the dissolution of MCl: $[\text{Cl}^-]_{total} = [\text{Cl}^-]_{initial} + [\text{Cl}^-]_{from MCl} = 0.1 + s$. Substitute these concentrations into the $K_{sp}$ expression: $K_{sp} = [\text{M}^+][\text{Cl}^-]_{total}$ $1 \times 10^{-10} = (s)(0.1 + s)$. Since MCl is sparingly soluble, its solubility $s$ in the presence of a common ion (0.1 M Cl$^-$) will be very small. Thus, we can assume $s \ll 0.1$. So, $0.1 + s \approx 0.1$. The equation becomes: $1 \times 10^{-10} \approx (s)(0.1)$. $1 \times 10^{-10} = 0.1s$. Solve for $s$: $s = \frac{1 \times 10^{-10}}{0.1} = \frac{1 \times 10^{-10}}{1 \times 10^{-1}} = 1 \times 10^{-10 - (-1)} = 1 \times 10^{-10+1} = 1 \times 10^{-9} \text{ M}$. Let's check the approximation: $s = 10^{-9}$. Is $10^{-9} \ll 0.1$? Yes, $10^{-9}$ is much smaller than $10^{-1}$. So the approximation is valid. The molar solubility of MCl in $0.1 \text{ M NaCl}$ solution is $1 \times 10^{-9} \text{ M}$. This matches option (c). \[ \boxed{10^{-9} \text{ M}} \]