Question

What is the dissociation constant of water (Kw) at 25oC? Calculate the pH of pure water at 25 oC (Show your work!). If the dissociation constant of water is increased at a different temperature, will the pH of water change? Explain your answer g

Answer 1

Explanation:

$H_2O+H_2O\rightleftharpoons H_3O^++OH^-$

The dissociation constant of the water is given as :

$K=\frac{[H_3O^+][OH^-]}{[H_2O][H_2O]}$

$K_w=K\times [H_2O][H_2O]=[H_3O^+][OH^-]$

The value of $K_w$ st 25°C = $1\times 10^{-14}$

$K_w=[H_3O^+][OH^-]$

$1\times 10^{-14}=[H_3O^+][OH^-]$

Water is neutral in nature with equal amount of hydronium and hydroxide ions.

$1\times 10^{-14}=[H_3O^+][H_3O^+]$

$[H_3O^+]=10^{-7} M$

$pH=-\log[H_3O^+]$

$pH=-\log[10^{-7} M]=7$

The pH of the water at 25°C is 7.

With increase in temperature dissociation of water will increase with which concentration of hydronium and hydroxide ions will also increase which will in return will lower the down the pH of the water at higher temperature.