10^{-6} M NaOH is diluted 100 times. The pH | vedclass

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(A) Initial concentration of $[OH^-] = 10^{-6} \ M$. Upon dilution by $100$ times,the concentration becomes $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \

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between $7$ and $8$

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(A) Initial concentration of $[OH^-] = 10^{-6} \ M$. Upon dilution by $100$ times,the concentration becomes $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \ M$. Since this concentration is very low,we must consider the contribution of $[OH^-]$ from the auto-ionization of water,which is $10^{-7} \ M$. Total $[OH^-] = 10^{-8} + 10^{-7} = 10^{-8} (1 + 10) = 11 	imes 10^{-8} \ M$. $pOH = -\log(11 	imes 10^{-8}) = 8 - \log(11) \approx 8 - 1.0414 = 6.9586$. $pH = 14 - pOH = 14 - 6.9586 = 7.0414$. Thus,the $pH$ lies between $7$ and $8$.

$10^{-6} \ M \ NaOH$ is diluted $100$ times. The $pH$ of the diluted base is

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