Calculate the pH of 0.02 mol dm-3 of Calcium hydroxide?

1 Answer
Al E.
May 13, 2018

Consider the dissociation of #Ca(OH)_2# in solution,

#Ca(OH)_2 rightleftharpoons Ca^(2+) + 2OH^(-)#

Moreover, let's assume this is a strong base that fully dissociates. To be sure, one stoichiometric equivalent of hydroxide ions will dissociate in proportion to the concentration of the salt.

Hence,

#[OH^-] = (0.02"mol")/"L" * (2OH^-)/(Ca(OH)_2) approx 0.04"M"#

and by extension,

#"pH" = 14 + log[OH^-] approx 12.60#

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