I am trying to finish my last lab of the semester but am having trouble figuring

Question

I am trying to finish my last lab of the semester but am having trouble figuring it out. Please explain how you get your answer!!

Data:

Cuvette 1: 4mL of CuSO4 and 6mL of HNO3

Cuvette 2: 7mL of CuSO4 and 3mL of HNO3

Cuvette 3: 10mL of CuSO4 and 0mL of HNO3 Cu2+ Concentration is 0.24M

The dilution factor for cuvette 1 is 0.4 and for cuvette 2 it's 0.7.

1. Calculate the actual concentration of Cu2+ ion in your samples(both samples). Because you are simply diluting a copper containing solution of known concentration, you can use the equation

(conc. of original solution )*(volume of standard Cu2+)=(concentration of dilute solution)*(total solution volume)

Explanation / Answer

C1 * V1 = C2 * V2  

therefore concentration in curvettte 1 and curvette 2 can be found out from the concentration in curvette 3

Cuvette 1 :

C1 * 4 = 0.24 * 10

C1 = 0.6 M

Cuvette 2

C2 * 7 = 0.24 * 10

C2 = 0.34 M

We know

D = Dilution Factor
    D1 = 0.4    D2=0.7
V = Volume

C = Concentration

old = Prior to dilution step

new = Following dilution step

D = C old/ C new

for

Cuvette 1 :: C new = C old / D = C1 / D1 = 0.15 M

Cuvette 2 :: C new = C old / D = C2 / D2 = 0.048 M

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