Question

Ka for H2O with Kw = 10^-14​

Answer 1

Answer:

kw =ka×kb.

=10^-7×10^-7

=10^-14

ka=10^-14÷10^-7

=10^-7.

Answer 2

The value of $K_a$ for water is $10^{-14}$

Explanation:

We are given:

$K_w$ for water = $10^{-14}$

Equation for the ionic product of water is given as, $K_w=[H^+]\times [OH^-]$

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

$2[H^+]=10^{-14}$

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

The chemical equation for the dissociation of water follows:

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

The expression of $K_a$ for above equation follows:

$K_a=\frac{[H^+][OH^-]}{[H_2O]}$

Concentration of pure solids and liquids are taken as 1 in the equilibrium constant expression. So, $[H_2O]=1$

Putting values in above expression, we get:

$K_a=[H^+]\times [OH^-]\\\\K_a=(10^{-7})\times (10^{-7})\\\\K_a=10^{-14}$

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