KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf

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KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14 [H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 = 10 pH+pOH=14 pOh = 14-pH = 14-10=4 Solution KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14 [H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 = 10 pH+pOH=14 pOh = 14-pH = 14-10=4.

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