Question

what is the degree of dissociation of water at 25°c?

Answer 1

The degree of dissociation of water at 25°C is 1.9*10^-7 % and density is 1g-cm^-3.

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Answer 2

55.55 M is Degree of Dissociation

Explanation:

Water generally behaves as a weak electrolyte and degree of dissociation at 298 k is defined by

$H_2O\ (aq)$ ⇔ $H^+\ (aq)$ + $OH^-\ (aq)$

Hence, the ionization constant of water can be written as

$K(H_2O) [H_2O] = [H^+][OH^-]$

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

The concentration of water is very high and a small fraction of it undergoes ionization, therefore the concentration of water [$H_2O$] may be taken as constant and combine with $K_{\alpha}(H_2O)$ The sum of $K_{\alpha}(H_2O)$ and $H_2O$ is called an ionic product of water, represented as $K_w.K_{\alpha}(H_2O) [H_2O] = [H^+] [OH^-]$

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

Experimentally at 298 K temperature, the concentration of $H^+$ ion in pure water is found to be $1.0 \times 10^{-7} molL^-^1$. Since in pure water, the concentration of $H^+$ ion is equal to the concentration of $OH^-$ ions as the dissociation of water produces an equal number of $H^+$ ions, therefore

$[H^+] = [OH^-] = 1.0 \times 10^{-7}$

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

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

$K_w= [1.0 \times 10^{-7}\ molL^-^1] [1.0 \times 10^{-7}\ molL^-^1]=1.0 \times 10^{-14} \frac {mol^2}{L^2}$

Since the concentration of $H^+\ and\ OH^-$ ions in pure water is $10^{-7}$ at 298 K temperature and [$H_2O$] is 55.55 M.