The concentration of NaOH solution is 10^{-8} | vedclass

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(D) For a very dilute solution of a strong base like $NaOH$ with concentration $10^{-8} \ M$,the contribution of $[OH^{-}]$ from the auto-ionization o

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Lies between $10^{-6} \ M$ and $10^{-7} \ M$

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(D) For a very dilute solution of a strong base like $NaOH$ with concentration $10^{-8} \ M$,the contribution of $[OH^{-}]$ from the auto-ionization of water cannot be neglected.$H_2O ightleftharpoons H^{+} + OH^{-}$,where $[OH^{-}]_{water} = 10^{-7} \ M$.The total concentration of $[OH^{-}]$ is the sum of $[OH^{-}]$ from $NaOH$ $(10^{-8} \ M)$ and $[OH^{-}]$ from water $(10^{-7} \ M)$.$[OH^{-}]_{total} = [OH^{-}]_{NaOH} + [OH^{-}]_{water} = 10^{-8} + 10^{-7} = 10^{-8} + 10 	imes 10^{-8} = 11 	imes 10^{-8} = 1.1 	imes 10^{-7} \ M$.This value is slightly greater than $10^{-7} \ M$ but significantly less than $10^{-6} \ M$.However,considering the options provided,the concentration is effectively dominated by the water contribution,making it closer to $10^{-7} \ M$. Given the options,the most accurate description is that it lies between $10^{-6} \ M$ and $10^{-7} \ M$.

The concentration of $NaOH$ solution is $10^{-8} \ M$. Find out the $[OH^{-}]$ concentration.

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