Question

340.0 gallons of 15 M nitric acid were added to a lake. The initial pH of the lake was 6.20 and the final pH was 4.60. If none of the acid was consumed in chemical reactions, determine the volume of the lake.

Answer 1

1gallon = 3.785 L

360 gallons = 340 x 3.785 = 1286.9 L

V1 = 1286.9 L

M1 = 15 M

lake initial pH = 6.20

[H+] = 10^-PH

          = 10^-.6.20

M2 = 6.31 x 10^-7 M

volume   of lake = V2 Litres

afte mixing HNO3 + lake water pH = 3.70

                        [H+] result = 10^-3.70 = 2 x 10^-4 M

now mixing molarity

[H+]results = (M1 V1 + M2 V2 ) / V1 + V2

2 x 10^-4 =   ( 15 x 1286.9 + 6.31 x 10^-7 x V2 ) / (1286.9 + V2 )

0.27252 + 2 x 10^-4 V2 = 19303 + 6.31 x 10^-7 x V2

20438.7 = 1.99 x 10^-4 V2

V2 = 1.02 x 10^8 L

volume of Lake = 1.02 x 10^8 L

   or

volume of Lake = 2.71 x 10^7 gallons