work out the [OH-], [H+] and pH of the solution formed when 0.500L of 0.250 M hydrochloric acid solution is mixed with 0.500 L of a 0.250 M calcium hydroxide solution.
I think it is easier to set up an ICE chart for these. Note that it is a limiting reagent problem.
mol HCl = M x L = 0.5 x 0.25 = 0.125
mol Ca(OH)2 = 0.5 x 0.25 = 0.125
.......2HCl + Ca(OH)2 ==> CaCl2 + 2H2O
I.....0.125....0.125......0........0
C....-0.125...-0.0625...0.0625..0.0625
E......0.......0.0625...0.0625..0.0625
Therefore, the final solution is one in which you have 0.0625 mol Ca(OH)2 in 1L of solution.
You can do OH, pOH, pH from there?