The pH of a solution produced when an aqueous | vedclass

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Question

(A) For the first solution,$pH = 6$,so $[H^{+}]_1 = 10^{-6} \ M$.For the second solution,$pH = 3$,so $[H^{+}]_2 = 10^{-3} \ M$.When equal volumes $(V)

Vedclass · Accepted Answer

$3.3$

Vedclass · Answer

(A) For the first solution,$pH = 6$,so $[H^{+}]_1 = 10^{-6} \ M$.For the second solution,$pH = 3$,so $[H^{+}]_2 = 10^{-3} \ M$.When equal volumes $(V)$ are mixed,the total volume becomes $2V$.The total concentration of $[H^{+}]$ is given by: $[H^{+}]_{total} = \frac{[H^{+}]_1 V + [H^{+}]_2 V}{2V} = \frac{10^{-6} + 10^{-3}}{2}$.$[H^{+}]_{total} = \frac{0.001001}{2} = 5.005 	imes 10^{-4} \ M$.The $pH$ is calculated as: $pH = -\log(5.005 	imes 10^{-4}) = 4 - \log(5.005) \approx 4 - 0.7 = 3.3$.

The $pH$ of a solution produced when an aqueous solution of $pH$ $6$ is mixed with an equal volume of an aqueous solution of $pH$ $3$ is about:

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