A solution has pH = 5. If it is diluted 100 t | vedclass

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(A) Initial concentration of $[H^{+}] = 10^{-5} \ M$.After diluting $100$ times,the concentration of $[H^{+}]$ becomes $10^{-5} / 100 = 10^{-7} \ M$.S

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Neutral

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(A) Initial concentration of $[H^{+}] = 10^{-5} \ M$.After diluting $100$ times,the concentration of $[H^{+}]$ becomes $10^{-5} / 100 = 10^{-7} \ M$.Since the concentration is very low,the $[H^{+}]$ contribution from water $(10^{-7} \ M)$ cannot be neglected.Total $[H^{+}] = 10^{-7} + 10^{-7} = 2 	imes 10^{-7} \ M$.$pH = -\log(2 	imes 10^{-7}) = 7 - \log(2) = 7 - 0.3010 = 6.699$.Since the $pH$ is slightly less than $7$,the solution remains slightly acidic,but it is often considered nearly neutral in many contexts. However,among the given options,it is closest to neutral.

$A$ solution has $pH = 5$. If it is diluted $100$ times,what will be the nature of the resulting solution?

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