Question

Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 × 10−13 at some temperature, compute the pH of a neutral aqueous solution at that temperature.

Answer 1

Since Kw= [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²
So,
3.31x10⁻¹³ = [H⁺]²
Take the square root= 5.75x10⁻⁷
Then take the negative log to find the pH:
-log(5.75x10⁻⁷) = 6.25

Answer 2

The pH of the aqueous solution is 6.2.

The equilibrium constant Kw describes the dissociation of water into hydrogen and hydroxide ions. Like every other equilibrium constant, it depends on temperature.

H2O(aq) -----> [OH]^-(aq) + [H]^+(aq)

If we know that Kw = 3.31 × 10^−13

And

Kw =  [OH]^-  [H]^+

We know that;

[OH]^- = [H]^+ = ([H]^+)^2

Kw = ([H]^+)^2

[H]^+ = √Kw

[H]^+ = √(3.31 × 10^−13)

[H]^+ =5.75 × 10^−7

Since;

pH = - log[H]^+

pH = -log (5.75 × 10^−7)

pH = 6.2

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