Fill the cuvettes with the following amounts of 0.06 M copper (II) sulfate and 1

Question

Fill the cuvettes with the following amounts of 0.06 M copper (II) sulfate and 1 M nitric acid:

CuSO4 (mL)                HNO3 (mL)             Cu+2 Concentration (mol/L)
cuvette 1: 4                6                               ?
cuvette 2: 7                3                               ?
cuvette 3: 10              0                               0.06 M

Use the dilution formula, M1V1 = M2V2, to calculate the concentration of Cu+2 ions in cuvettes 1 and 2

I got .04 and .14, but another answer on Chegg says it's .024 and .042. Please help, thank you.

Explanation / Answer

cuvette 1 :

volume of CuSO4 = 4 mL

concentration of Cu+2 = 0.06 M

volume of HNO3 = 6 mL

final volume of CuSO4 = 4 + 6 = 10 mL

M1 V1 = M2 V2

4 x 0.06 = M x 10

M = 0.024

concentration of Cu+2 = 0.024 M

for cuvvette 2 :

V1 = 7 mL

V2 = 7 + 3 = 10 mL

M1 = 0.06 M

7 x 0.06 = M x 10

M = 0.042

concentration of Cu+2 = 0.042 M

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