Question

At 25c the dissociation constant for pure water is given by ?

1.(55.5×10)^-1

2.(1×10)^-14

3.(1×10^-14)/18

4. None of these

Answer 1

Answer : The correct option is, (2) $1\times 10^{-14}$

Explanation :

Dissociation constant of pure water : It is defined as the product of the concentration of hydrogen ion and hydroxide ion.

The equilibrium dissociation reaction for water will be:

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

The expression for dissociation constant for water will be:

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

Experimentally, at $25^oC$ temperature, the concentration of $H^+$ ion in pure water is found to be $1.0\times 10^{-7}mol/L$.

As we know that in pure water, the concentration of hydrogen ion is equal to the concentration of hydroxide ions.

So, the value of dissociation constant of pure water will be:

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

$k_w=(1.0\times 10^{-7})\times (1.0\times 10^{-7})$

$k_w=(1.0\times 10^{-14})$

Therefore, the dissociation constant of pure water is by $(1.0\times 10^{-14})$

Answer 2

Answer:

Explanation:

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