Calculate the pH of a 0.0131 mol dm^-3 solution of calcium hydroxide at 10 degrees centigrade.
Multiply by 2 because calcium hydroxide = Ca(OH)2 so 2 x [OH-] per molecule.[OH-] = 0.0131 x 2 = 0.0262
Insert [OH-] value into the equilibrium equation along with the value of Kw at 10 degrees centigrade 2.93 x 10^-15 (from data tables). This gives the value for [H=].[H+] = (Kw/[OH-] ) = 2.93 x 10^-15 / 0.0262 = 1.118 x 10^-13
Finally, insert [H+] value into pH equation.pH = -log (1.118 x 10^-13) = 12.95
EW
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