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

1 Answer
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