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(A) Let the volume of each solution be $V \ L$. For the first solution,$[H^+]_1 = 10^{-6} \ M$. For the second solution,$[H^+]_2 = 10^{-3} \ M$. When

Vedclass · Accepted Answer

$3.3$

Vedclass · Answer

(A) Let the volume of each solution be $V \ L$. For the first solution,$[H^+]_1 = 10^{-6} \ M$. For the second solution,$[H^+]_2 = 10^{-3} \ M$. When mixed,the total volume becomes $2V \ L$. The total moles of $H^+$ ions = $(10^{-6} 	imes V) + (10^{-3} 	imes V) = V(10^{-6} + 10^{-3}) \ mol$. The concentration of the resulting solution is $[H^+]_{mix} = \frac{V(10^{-6} + 10^{-3})}{2V} = \frac{10^{-3} + 0.001 	imes 10^{-3}}{2} \approx \frac{10^{-3}}{2} = 0.5 	imes 10^{-3} = 5 	imes 10^{-4} \ M$. Now,$pH = -\log[H^+]_{mix} = -\log(5 	imes 10^{-4}) = -(\log 5 + \log 10^{-4}) = -(0.7 - 4) = 3.3$.

When equal volumes of solutions with $pH = 6$ and $pH = 3$ are mixed,what is the $pH$ of the resulting solution?

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