A solution with pH = 5.0 is diluted 100 times | vedclass

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(D) The initial concentration of $H^+$ ions is $[H^+]_1 = 10^{-pH} = 10^{-5} \ M$. Upon diluting the solution $100$ times,the new concentration become

Vedclass · Accepted Answer

$6.96$

Vedclass · Answer

(D) The initial concentration of $H^+$ ions is $[H^+]_1 = 10^{-pH} = 10^{-5} \ M$. Upon diluting the solution $100$ times,the new concentration becomes $[H^+]_2 = \frac{10^{-5}}{100} = 10^{-7} \ M$. Since the concentration of $H^+$ from water $(10^{-7} \ M)$ cannot be ignored when the acid concentration is very low,the total $[H^+]$ is $[H^+]_{total} = [H^+]_{acid} + [H^+]_{water} = 10^{-7} + 10^{-7} = 2 	imes 10^{-7} \ M$. The $pH$ is calculated as $pH = -\log[H^+]_{total} = -\log(2 	imes 10^{-7}) = 7 - \log 2 = 7 - 0.301 = 6.699 \approx 6.7$.

$A$ solution with $pH = 5.0$ is diluted $100$ times. Calculate the $pH$ of the diluted solution.

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