The pH of a H_2SO_4 solution is 5.0. If more | vedclass

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Question

(C) The relationship between $pH$ and $H^{+}$ concentration is given by the formula: $pH = -\log[H^{+}]$.Initially,$pH_1 = 5.0$,so $[H^{+}]_1 = 10^{-5

Vedclass · Accepted Answer

$1000$

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(C) The relationship between $pH$ and $H^{+}$ concentration is given by the formula: $pH = -\log[H^{+}]$.Initially,$pH_1 = 5.0$,so $[H^{+}]_1 = 10^{-5} \ M$.Finally,$pH_2 = 2.0$,so $[H^{+}]_2 = 10^{-2} \ M$.The ratio of the final concentration to the initial concentration is: $\frac{[H^{+}]_2}{[H^{+}]_1} = \frac{10^{-2}}{10^{-5}} = 10^{3} = 1000$.Therefore,the concentration of $H^{+}$ ions becomes $1000$ times the initial concentration.

The $pH$ of a $H_2SO_4$ solution is $5.0$. If more $H_2SO_4$ is added to this solution,the $pH$ becomes $2.0$. The concentration of $H^{+}$ ions will become how many times the initial concentration?

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