hitunglah pH larutan natrium fenolat NaOC6H5, 1,O x 10 pangkat -3 M? Kb untuk HOC6H5 ialah 1.05 x10 pangkat-10?

[OH-] = akar M x kb = akar 1,0x10^-3 . 1.05 x 10^-10= akar 1,05 x 10^-12 = akar 0.105 x 10^-12 = 0,324 x 10^-6 = 3,24x10^-7
pOH = -log [OH-] = -log 3,24x10^-7 = 7-log3,24
pH = 14-pOH = 14-(7-log3,24) = 7-log3,24