Mennyi annak a nátrium-acetát-oldatnak a koncentrációja mol/dm3-ben, amelynek a pH-ja 8,73? Ks (ecetsav) = 1,9.10–5 mol / dm3

pH=14-pOH

Kb=[OH-]^2/(c-[OH-])

pOH=14-pH=14-pH

[OH-]=10^-pOH=10^-(14-pH)

Kb=[OH-]^2/(c-[OH-])=[10^-(14-pH)]^2/[c-10^-(14-pH)]

Kb*[c-10^-(14-pH)]=[10^-(14-pH)]^2

Kb*c-Kb*10^-(14-pH)=[10^-(14-pH)]^2

Kb*c=[10^-(14-pH)]^2+Kb*10^-(14-pH)

c={[10^-(14-pH)]^2+Kb*10^-(14-pH)}/Kb

c={[10^-(14-8.73)]^2+1.9*10^-5*10^-(14-8.73)}/(1.9*10^-5)=6,89*10^-6 M