Question

If pure water has PKw=13.36 at 50 Celsius ,the ph of pure water will be

Answer 1

Ionic product of water is the product of the molar concentrations of $H_{3}O^{+}$ and $OH^{-}$.

$pK_{W} = -logK_{W}$

$13.36 = - log[K_{W}]$

$K_{W} = 10^{-13.36}$

$K_{W} = [H_{3}O^{+}][OH^{-}]$

$10^{-13.36}$ = $[H_{3}O^{+}][OH^{-}]$

The equilibrium representing the auto dissociation of water:

$2H_{2}O(l) <==> H_{3}O^{+}(aq) + OH^{-}(aq)$

At equilibrium, x = $[H_{3}O^{+}]=[OH^{-}]$

$10^{-13.36}$ = x(x)

x = $2.09 * 10^{-7} M$

$[H_{3}O^{+}] = 2.09*10^{-7} M$

pH = - log (2.09*$10^{-7}$

= 6.68

Answer 2

Answer:6.68

Explanation:

pH = pKw/2

Therefore pH = 13.36 /2 = 6.68