Answer
$pH \approx 11.472$
Work Step by Step
1. Convert $10^{-6}g$ of $Mn^{2+}$ to number of moles:
$mm(Mn) = 54.94 g/mol$
$n(moles) = \frac{mass(g)}{mm}$
$n(moles) = \frac{10^{-6}}{54.94}$
$n(moles) = 1.82 \times 10^{-8}$
$10^{-6} g/L (Mn)= 1.82 \times 10^{-8} M$
2. Use the Ksp formula to find $[OH^-]$
$Ksp = [Mn^{2+}][OH^-]^2$
$1.6 \times 10^{-13} = [1.82 \times 10^{-8}][OH^-]^2$
$8.79 \times 10^{-6} = [OH^-]^2$
$[OH^-] = 2.965 \times 10^{-3}$
3. Find the pOH:
$pOH = -log[OH^-]$
$pOH = -log(2.965 \times 10^{-3})$
$pOH = 2.528$
4. Calculate the pH:
$pH = 14 - pOH$
$pH = 14 - 2.528$
$pH = 11.472$
