Question

340.0  gallons of 15.0 M nitric acid were added to a lake. The initial pH of the lake was 6.40 and the final pH was 4.50 . If none of the acid was consumed in chemical reactions, determine the volume of the lake. (IN LITERS)

Answer 1

concentration of nitric acid is 15 M

volume of nitric acid is 340 gallons.

1 L = 0.264172 gallons

340 gallons = 340 x ( 1 L / 0.264172) = 1287 L

convert the volume into liters by using the following conversion

calculate the number of moles of acid as shown below

Number of moles = concentration x volume

= 15 M x 1287 L = 19305 mol

initial pH of the lake is 6.40

calculate the initial concentration of H+ in the lake as shown below

pH = -log[H⁺]

[H⁺] = 10^-pH

= 10^-6.40 = 3.98 x 10^-7 M

final pH of the lake is 4.50

final concentration of H⁺ , [H⁺] = 10^-pH

[H⁺] = 10^-4.50 = 3.1622 x 10^-5 M

change in concentration H⁺ in the lake = 3.1622 x 10^-5 M - 3.98 x 10^-7 M = 3.1224 x 10^-5 M

calculate the volume of the lake as shown below.

volume = number of moles / concentration

= 19305 mol / (3.1224 x 10^-5 M) = 618274404.3 L

therefore the volume of the lake is 618274404.3 L

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