Both HCl and Ca(OH)_2 are strong electrolytes, so each will ionize 100% in solution.
This makes it possible to determine the amount of H^+ in the HCl solution and the amount of OH^- in the Ca(OH)_2 solution, by multiplying concentration x volume in each case:
HCl: (0.250 "mol"/L)(0.500 L) = 0.125 "mol" of HCl
and since it is strong, moles H^+=0.125 as well.
Ca(OH)_2: (0.250 "mol"/L)(0.500 L) = 0.125 "mol"
and since it is strong and because each one mole of Ca(OH)_2 contains two moles of OH^- ions, moles OH^-=0.250.
Looking at neutralization, we see that the reaction is one-to-one between H^+ and OH^-
H^+ + OH^- rarr H_2O
So, the 0.125 mol of H^+ consumes an equal amount of OH^-, but 0.125 mol OH^- will remain, now in a total volume of 1.0 L
Therefore
[OH^-] = 0.125 "mol"/ 1.0 L = 0.125 M
Since [OH^-] = 0.125 M,
[H^+] = (1xx10^(-14))/0.125 = 8 xx 10^(-14) M
and the pH is #-log(8 xx 10^(-14)) = 13.1
