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Auto Protolysis of Water

Question

The ionic product of water is $10^{- 14}$ , What is the $H^{+}$ ion concentration of a 0.1 M Na OH solution?
14M
$10^{- 13} M$
$10^{- 14} M$
13M

A

10^{- 13} M

B

14M

C

10^{- 14} M

D

13M

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Solution

Verified by Toppr

NaOH is a strong electrolyte and completely dissociates in solution.
Hence, $[O H^{-}] = 0.1$ .
The ionic product of water is $10^{- 14}$ .
Hence, $k_{w} = [H^{+}] [O H^{-}] = 10^{- 14}$ or $[H^{+}] = \frac{10^{- 14}}{[O H^{-}]} = \frac{10^{- 14}}{0.1} = 10^{- 13}$ .

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