An acid solution of pH=6 is diluted 1000 time | vedclass

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(D) Given,$pH = 6$. Therefore,the initial concentration of $[H^{+}] = 10^{-6} \ M$. After diluting the solution $1000$ times,the new concentration of

Vedclass · Accepted Answer

$6.99$

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(D) Given,$pH = 6$. Therefore,the initial concentration of $[H^{+}] = 10^{-6} \ M$. After diluting the solution $1000$ times,the new concentration of $[H^{+}]$ from the acid is $[H^{+}]_{acid} = \frac{10^{-6}}{1000} = 10^{-9} \ M$. Since this concentration is very low,the contribution of $[H^{+}]$ from the auto-ionization of water $(10^{-7} \ M)$ cannot be neglected. Total $[H^{+}] = [H^{+}]_{acid} + [H^{+}]_{water} = 10^{-9} + 10^{-7} \ M$. Total $[H^{+}] = 10^{-7} (0.01 + 1) = 1.01 	imes 10^{-7} \ M$. $pH = -\log(1.01 	imes 10^{-7}) = 7 - \log(1.01) \approx 7 - 0.0043 = 6.9957$. Thus,the $pH$ of the final solution is approximately $6.99$.

An acid solution of $pH=6$ is diluted $1000$ times,the $pH$ of the final solution becomes

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