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