Solubility Question Help Please

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The K of Mg(OH) in water is 1.2 × 10–11 sp 2

mol3/L3. If the Mg2+ concentration in an acid solution is 1.2 × 10–5, what is the pH at which Mg(OH)2 just begins to precipitate?

A.3 B.4 C.5 D. 11

The answer is D (11) but I do not understand the explanation. How would you do this question in a timely manner if presented on MCAT?

 

[theonlytycrane]

Mg(OH)2 <-> Mg(2+) + 2(-OH). [Mg2+] is given as 1.2 * 10^-5. so [-OH] is twice that = 2.4 * 10^-5.

pOH = -log10(2.4 * 10^-5) = 5 - log(2.4) using the approximation -log10(a * 10^-b) = b - log(a).

pOH + pH = 14. pH = 14 - (5 - log(2.4)) which is closest to answer (D).