in HCl [H+] concentration = 0.12M pH of HCl = -log(o.12) =0.92 in Ca(OH)2 [OH-] concentration = 2*0.15 =0.3M pOH = -log([OH-] PH = 14-pH pH of Ca(OH)2 = 14+log(2*0.15) =13.477 we know Kw of water =10^-14 [H+][OH-] =10^-14 so both have equalconcentrations hence [H+] = [OH-] = 10^-7 M pH of water = -log(10^-7) = 7 Solution in HCl [H+] concentration = 0.12M pH of HCl = -log(o.12) =0.92 in Ca(OH)2 [OH-] concentration = 2*0.15 =0.3M pOH = -log([OH-] PH = 14-pH pH of Ca(OH)2 = 14+log(2*0.15) =13.477 we know Kw of water =10^-14 [H+][OH-] =10^-14 so both have equalconcentrations hence [H+] = [OH-] = 10^-7 M pH of water = -log(10^-7) = 7.