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
