The ionic product of water, Kw = 2.93 × 10−15 mol dm−6 at 10 °C. Calculate the pH of a 0.0131 mol dm−3 solution of calcium hydroxide at 10 °C Give your answer to two decimal places.
[OH-] = 0.0262 mol dm-3[H+] = (Kw/[OH-]) = 2.93 x 10−15 / 0.0262 (= 1.118 x 10−13) pH = (− log (1.118 x 10−13) = 12.9514 = 12.95
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