Question

At 40(degrees), the ion-product constant of water (Kw) is 2.00x10^-14. what is the ph of the pure water at this temperature ?

Answer 1

We can assume that the concentrations of hydronium and hydroxide are equal, so:\[K_w=[H_3O^+][OH^-]=2.00 \times {10^{-14}}=(x)(x)=x^2\]\[x=1.41 \times {10^{-7}}M\]\[pH=-\log([H^+])=6.85\]