Question

At \( T(K) \), the solubility product of \( AX \) is \( 10^{-10} \). What is the molar solubility of \( AX \) in 0.1 M HX solution?

• A. \( 10^{-5} \)
• B. \( 10^{-10} \)
• C. \( 10^{-9} \)
• D. \( 10^{-8} \)

Hint:

When the solubility product is affected by the presence of a common ion, approximate the concentration of the common ion to be constant and solve the equation accordingly.

Answer 1

The Correct Option is C

We are given the solubility product of \( AX \) as \( K_{sp} = 10^{-10} \) at \( T(K) \). Let the molar solubility of \( AX \) be \( s \) in pure water. So, the dissociation of \( AX \) is represented as: \[ AX (s) \rightleftharpoons A^+ (aq) + X^- (aq), \] where the concentration of \( A^+ \) and \( X^- \) will both be \( s \) in pure water. Therefore, the solubility product is: \[ K_{sp} = [A^+][X^-] = s^2. \] Thus, the solubility \( s \) in pure water is: \[ s = \sqrt{K_{sp}} = \sqrt{10^{-10}} = 10^{-5} \, {mol/L}. \] Now, in the presence of \( 0.1 \, {M} \) HX, the concentration of \( X^- \) ions is increased by the dissociation of HX. Since HX dissociates fully: \[ {[X}^-] = 0.1 \, {M} + s. \] Now, the solubility product equation becomes: \[ K_{sp} = [A^+][X^-] = s(0.1 + s). \] Since \( s \) is much smaller than 0.1, we can approximate \( 0.1 + s \approx 0.1 \). Therefore, the equation simplifies to: \[ K_{sp} = s \cdot 0.1. \] Substituting the value of \( K_{sp} = 10^{-10} \): \[ 10^{-10} = s \cdot 0.1, \] \[ s = \frac{10^{-10}}{0.1} = 10^{-9} \, {mol/L}. \] Thus, the molar solubility of \( AX \) in 0.1 M HX solution is \( 10^{-9} \, {mol/L} \).