If 100 , mL of a solution with pH = 3 is mixe | vedclass

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Question

(B) The $pH$ of a solution is defined as $pH = -\log[H^+]$.For the first solution,$pH = 3$,so $[H^+]_1 = 10^{-3} \, M$.For the second solution,$pH = 3

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The $pH$ of a solution is defined as $pH = -\log[H^+]$.For the first solution,$pH = 3$,so $[H^+]_1 = 10^{-3} \, M$.For the second solution,$pH = 3$,so $[H^+]_2 = 10^{-3} \, M$.When two solutions of the same substance with the same $pH$ are mixed,the concentration of $[H^+]$ ions in the mixture remains the same.$[H^+]_{mix} = \frac{V_1[H^+]_1 + V_2[H^+]_2}{V_1 + V_2} = \frac{100 	imes 10^{-3} + 400 	imes 10^{-3}}{100 + 400} = \frac{500 	imes 10^{-3}}{500} = 10^{-3} \, M$.Therefore,the $pH$ of the mixture is $-\log(10^{-3}) = 3$.

If $100 \, mL$ of a solution with $pH = 3$ is mixed with $400 \, mL$ of another solution with $pH = 3$,what will be the $pH$ of the resulting mixture?

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