#### Question

Ionic product of water at 310 K is 2.7 × 10^{–14}. What is the pH of neutral water at this temperature?

#### Solution

Ionic product,

`K_w = [H^+] [OH^-]`

Let `[H^+] = x`

Since `[H^+] = [OH^(-)], K_w = x^2`

`=> `K_w` "at 310K" is 2.7 xx 10^(-14)`

`:. 2.7 xx 10^(-14) = x^2`

`=> x = 1.64 xx 10^(-7)`

`=> [H^+] = 1.64 xx 10^(-7)`

`=> pH = -log[H^+]`

`= -log [1.64 xx 10^(-7)]`

= 6.78

Hence, the pH of neutral water is 6.78

Is there an error in this question or solution?

Solution Ionic Product of Water at 310 K is 2.7 × 10–14. What is the Ph of Neutral Water at this Temperature? Concept: Ionization of Acids and Bases - The Ionization Constant of Water and Its Ionic Product.