The Ksp for magnesium hydroxide (Mg(OH)2) is 1.1 x 10-11. What is the molar solubility of the magnesium ion if the pH is 10.0?

Question

rogue · Answer

pH + pOH = 14
pH = 10 ---> pOH = 4
[OH-] = 10^(-pOH)
[OH-] = 10^-4 M

$$Mg(OH)_2 (s) \rightarrow Mg^{2+} (aq) + 2OH^- (aq)$$$$K_{sp} = [Mg^{2+}][OH^-]^2$$$$1.1 \times 10^{-11} = [Mg^{2+}] \times (10^{-4})^2$$$$[Mg^{2+}] = \frac {1.1 \times 10^{-11}}{10^{-8}} = 1.1 \times 10^{-3} M$$