When a 10^{-6} M NaOH solution is diluted 1 | vedclass

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Question

(C) Initial concentration of $NaOH$ is $10^{-6} \ M$. After diluting $100$ times,the new concentration is $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \ M$

Vedclass · Accepted Answer

Between $7$ and $8$

Vedclass · Answer

(C) Initial concentration of $NaOH$ is $10^{-6} \ M$. After diluting $100$ times,the new concentration is $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \ M$. Since the concentration is very low,we must consider the contribution of $OH^-$ ions from water $(10^{-7} \ M)$. Total $[OH^-] = 10^{-8} + 10^{-7} = 10^{-8} + 10 	imes 10^{-8} = 11 	imes 10^{-8} \ M$. $pOH = -\log(11 	imes 10^{-8}) = 8 - \log(11) \approx 8 - 1.04 = 6.96$. $pH = 14 - pOH = 14 - 6.96 = 7.04$. Therefore,the $pH$ lies between $7$ and $8$.

When a $10^{-6} \ M \ NaOH$ solution is diluted $100$ times,the $pH$ of the diluted solution will be ............ .

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