Sebanyak 200 ml larutan Ca(OH)2 0,03 M mempunyai pH sebesar


Ca(OH)2 termasuk dalam basa kuat, sehingga OH- = 2 x 0,03 = 0,06 M = 6 x 10^-2, pOH = 2-Log6, pH = 14-(2-Log6), = 12+log6
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Dik : vol. Ca(OH)2 = 200 mL = 0,2 liter, Molaritas Ca(OH)2 = 0,03 M, Dit: pH = ….?, Jawab :, reaksi ionisasi :Ca(OH)2, OH^- = 2 x 0,03, OH^- = 6 x 10^-2, pOH = -logOH^-, pOH = – log 6 x 10^-2, pOH = 2- log 6, pOH = 2 – 0,448, pOH = 1,222, pH = 14 – pOH, pH = 14 – 1,222, pH = 12,778,