pH values of HCl and NaOH solutions each of s | vedclass

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(B) For $HCl$ (a strong acid): $[H^+] = \frac{N}{100} = 0.01 \ M = 10^{-2} \ M$. $pH = -\log[H^+] = -\log(10^{-2}) = 2$. For $NaOH$ (a strong base): $

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$2$ and $12$

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(B) For $HCl$ (a strong acid): $[H^+] = \frac{N}{100} = 0.01 \ M = 10^{-2} \ M$. $pH = -\log[H^+] = -\log(10^{-2}) = 2$. For $NaOH$ (a strong base): $[OH^-] = \frac{N}{100} = 0.01 \ M = 10^{-2} \ M$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. Since $pH + pOH = 14$ at $25^{\circ}C$,$pH = 14 - pOH = 14 - 2 = 12$. Therefore,the $pH$ values are $2$ and $12$ respectively.

$pH$ values of $HCl$ and $NaOH$ solutions each of strength $\frac{N}{100}$ will be respectively:

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