The pH of neutral water at $25^{\circ} \mathrm{C}$ is 7.0 . As the temperature increases, ionisation

Question

The pH of neutral water at $25^{\circ} \mathrm{C}$ is 7.0 . As the temperature increases, ionisation of water increases, however, the concentration of $\mathrm{H}^{+}$ions and $\mathrm{OH}^{-}$ions are equal. What will be the pH of pure water at $60^{\circ} \mathrm{C}$ ?

Professor PC Thomas Classes · Answer

(c) : The pH of neutral water at $25^{\circ} \mathrm{C}$ is 7.0 . $ \	herefore \quad\left[\mathrm{H}^{+}ight]=\left[\mathrm{OH}^{-}ight]=10^{-7} \ \quad\left(\because \mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]ight) $ \ Now, $K_w=\left[\mathrm{H}^{+}ight]\left[\mathrm{OH}^{-}ight]=\left(1 	imes 10^{-7}ight)^2=1 	imes 10^{-14}$ \ As the temperature increases, ionisation of \ water increases, thus [ $\mathrm{H}^{+}$] and [ $\mathrm{OH}^{-}$] increases equally. \ Now $ K_w=\left[\mathrm{H}^{+}ight]\left[\mathrm{OH}^{-}ight]>1 	imes 10^{-14} $ $ \left(\because\left[\mathrm{H}^{+}ight]=\left[\mathrm{OH}^{-}ight]ight) \ $ or $\left[\mathrm{H}^{+}ight]^2>1 	imes 10^{-14}$  $ 	herefore\left[\mathrm{H}^{+}ight]>1 	imes 10^{-7} 	ext { and } \mathrm{pH}<7 $