(C) Given: $[OH^{-}] = 10^{-9} \ M$ at $25^{\circ}C$.We know that $K_w = [H^{+}] [OH^{-}] = 10^{-14}$.Therefore,$[H^{+}] = \frac{10^{-14}}{10^{-9}} =

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(C) Given: $[OH^{-}] = 10^{-9} \ M$ at $25^{\circ}C$.We know that $K_w = [H^{+}] [OH^{-}] = 10^{-14}$.Therefore,$[H^{+}] = \frac{10^{-14}}{10^{-9}} = 10^{-5} \ M$.The $pH$ is calculated as $pH = -\log[H^{+}]$.$pH = -\log(10^{-5}) = 5$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Water hydrolysis and pH scale

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At $25^{\circ}C$,if $[OH^{-}] = 10^{-9} \ M$,find the $pH$ value of the solution.

A
$9$
B
$10$
C
$5$
D
$11$

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2 Equilibrium: Ionic Equilibrium

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Water hydrolysis and pH scale

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At $25^{\circ}C$,if $[OH^{-}] = 10^{-9} \ M$,find the $pH$ value of the solution.

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