[Ca(OH)2] = 0.0105M; therefore, 10 mL diluted to 500 must have a concn of
0.0105M x (10/500) = ??M
Since there are two OH ions per molecule of Ca(OH)2, (OH^-) must be twice that, then pOH = -log(OH^-).
pH + pOH = pKw = 14; use that to calculate H+.
Yes, that's correct but I wouldn't round it to 11 but keep it as 10.62. (10.62 has two significant figures --besides the 10 which comes from the log part--and you are allowed two--and I would keep 10.62.