410.0 gallons of 15 M nitric acid were added to a lake. The initial pH of the la

Question

410.0 gallons of 15 M nitric acid were added to a lake. The initial pH of the lake was 5.70 and the final initial pH was 4.10. If none of the acid was consumed in chemical reactions, determine the volume of the lake Analyze First calculate the moles of HNO_3 added to the lake, and then determine the increase in the concentration of H^+ in the lake from pH 5.700 to pH 4.1000. Knowing that, simply divide the moles of acid added by the molarity change of H^+ in the lake to obtain the size of the lake.

Explanation / Answer

1 gallon = 3.78541 L

410.0 gallon = 410.0 * 3.78541 L

= 1552 L

mol of HNO3 added = M*V

= 15 M * 1552 L

=23280 mol

initial pH = 5.70

use:

initial pH = - log [H+]initial

5.70 = - log [H+]initial

[H+]initial = 1.995*10^-6 M

final pH = 5.70

use:

final pH = - log [H+]final

4.10 = - log [H+]final

[H+]final = 7.943*10^-5 M

mol of H+ added = ([H+]final - [H+]initial)*volume of lake

(7.943*10^-5 - 1.995*10^-6 )*V

This must equal mol of HNO3 added.

(7.943*10^-5 - 1.995*10^-6 )*V = 23280

7.7435*10^-5 * V = 23280

V = 3.006*10^8 L

Answer: 3.006*10^8 L

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