HN7 448F17pd Word u ReferencesMailing Reviee View Nos Heading oding 2e Subiale N

Question

HN7 448F17pd Word u ReferencesMailing Reviee View Nos Heading oding 2e Subiale Normal 1 No SpHeading1 Heading 2 T Paragraph 1. A creek that receives runoff from an old mine enters a small river that drains a watershed with many other historic mining sites. The discharge of the creek is 0.75 cfs (ft3/sec) and that of the river is 4 cfs. a. If the cadmium concentration of the water in the creek is 345 gL and that of the river upstream from the confluence is 57 g. calculate the Cd concentration in the river downstream from the confluence with the creek. (5 pts) b. The pH of the river water is 5.43 and that of the creek water is 3.21. What pH would you predict to find in the river water downstream from the confluence? (4 p

Explanation / Answer

We have to calculate the sum of solids going in flow combined per unit volume:

1cfs = 28.316 l/s

0.5 cfs = 0.5 x 28.316 l/s = 14.258 l/s

4 cfs = 4 x 28.316 l/s = 113.264l/s

Total mass flow rate of solids = mass flow in Creek + mass flow in river = 14.258l/s x 345ug/l + 113.264 l/s x 57 ug/l

=11375 ug/s

Total volume flow rate = 14.258 l/s + 113.264 l/s = 127.522 l/s

Concentration after confluence= mass flow rate /volume flow rate =(11375 ug/s) / (127.522 l/s ) = 89.2ug/l

B)

Assuming this concentration to be dilute , we can add the amount of H+ ion concentration

pH = -log[H+]

5.43 = -log[H+]

[H+] = 3.715 x 10-6mol/l

Similarly [H+] of Creek = 6.16 x 10-4mol/l

moles of [H+] in river = 3.715 x 10-6mol/l x 113.264 l/s = 4.208 x 10-4mol/s

Moles of [H+] in creek = 6.16 x 10-4 mol/l x 14.258 l/s =8.783 x 10-3mol/s

New hydrogen ion concentration= total moles /total volume

7.217 x 10-5 mol/l

pH = -log[7.217 x 10-5mol/l] = 4.14

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