$K_{s p}$ of $Ca (OH) X_{2}$ is $4.0 \times 1 0^{- 6}$ . Taking $lo g 2 = 0.301$ , calculate the pH of its saturated solution at 298 K (two decimals).

Numerical answer type, tolerance 0.02.

Answer: 12.3

s = (K_{s p} /4)^{1/3} = 1.0 \times 1 0^{- 2} M

, so

[OH X^{-}] = 2 s = 2.0 \times 1 0^{- 2} M

pOH = 2 - 0.301 = 1.699

and pH

= 12.30

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