Ncert Exercises & Solutions

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

Ionic product,

K_{w} = [H^{+}] [{OH}^{-}]

Let [H^{+}] = x .

Since [H^{+}] = [{OH}^{-}], K_{w} = x^{2} .

\Rightarrow K_{w} at 310 K is 2.7 \times 10^{- 14} .

∴ 2.7 \times 10^{- 14} = x^{2}

\Rightarrow x = 1.64 \times 10^{- 7}

\Rightarrow [H^{+}] = 1.64 \times 10^{- 7}

\Rightarrow pH = - \log [H^{+}]

= - \log [1.64 \times 10^{- 7}]

= 6.78

Hence, the pH of neutral water is 6.78.