Question

The \( K_{sp} \) for Mn(OH)\(_2\) is \( 1.6 \times 10^{-13} \). What is the molar solubility of this compound in water?

• A. $\sqrt{40 \times 10^{-14}}$
• B. $1.6 \times 10^{-13}$
• C. $\sqrt{40 \times 10^{-13}}$
• D. $\sqrt{40 \times 10^{-14}}$

Hint:

To calculate molar solubility from Ksp, express the solubility expression in terms of the solubility (s) and solve for it.

Answer 1

The Correct Option is A

For Mn(OH)$_2$, the solubility product is given by: \[ K_{sp} = [Mn^{2+}][OH^-]^2 \] Assuming the molar solubility is $s$, the concentration of Mn$^{2+}$ is $s$ and the concentration of OH$^-$ is $2s$. Thus: \[ K_{sp} = s(2s)^2 = 4s^3 \] Solving for $s$: \[ 1.6 \times 10^{-13} = 4s^3 \quad \Rightarrow \quad s^3 = \frac{1.6 \times 10^{-13}}{4} = 4 \times 10^{-14} \] \[ s = \sqrt{40 \times 10^{-14}} = \sqrt{40} \times 10^{-7} { mol/L} \]