Question

Ionic product of water at 310k is 2.7 find ph of neutral water at this temperature

Answer 1

Ionic product { Kw } = [H3O+][OH-]

Kw = [H3O+][OH-] = 2.7 × 10^-14 at 310K

we know,

H2O+ H2O <=> [H3O+][OH-]

[H3O+] = [OH-]

therefore, [H3O+] = √{2.7 × 10^-14}

[H3O+] = 1.643 × 10^-7 M

PH = -log[H3O+] = -log(1.643 × 10^-7)

= -{log10^-7 + log(1.643)}

= -{ -7 + 0.2156}

=7 - 0.2156 = 6.7844

hence, PH = 6.7844