Question

Calculate [H3O+] and [OH−] for the solution

pH= 2.88

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answer 1

Given data, pH = 2.88

In aqueous medium proton doesnot exist free but it protonates water (H2O) molecule and forms H3O⁺ ions. The H+ and H3O+ ions are numerically same i.e [H3O+] = [H+]

pH is the -ve logarithm of H+ ion concentration.

Mathematically, pH = -log₁₀[H⁺]

hence, log₁₀[H⁺] = -pH

[H⁺] = 10^-pH.

[H⁺] = 10^-2.88.

[H⁺] = 1.32 x 10^-3 M

i.e. [H3O+] = 1.32 x 10-3 M.

Now the ionic product of water is given as,

Kw = [H+][OH-] = 1 x 10^-14.

[H+][OH-] = 1 x 10^-14.

1.32 x 10^-3 x [OH-] = 1 x 10^-14.

[OH-] = (1 x 10-14 ) / (1.32 x 10^-3).

[OH-] = 7.58 x 10^-12 M.

[H3O+] = 1.32 x 10^-3 , [OH-] = 7.58 x 10^-12.