this experiment examines the effect of varying the concentrations of solutions i
ID: 1032224 • Letter: T
Question
this experiment examines the effect of varying the concentrations of solutions in -2 M, 1.0 × 10-4 M 4. Part D of the anode and cathode compartments. Concentrations of 1.0 M, 1.0 x 10 will be examined. Wh concentrations should be prepared. For each of the desired solutions calculate the amount of DI water to be added to the given stock solution. ile the 1.0 M concentrations are readily available, the other two on the table below, TABLE 9.1 DI Water to Add to Stock Solution (mL) Desired Solution 1.0 x 102 M CuSO, 1.0 x 10-4 M Cuso, 1.0 × 10-2 M ZnSO. 1.0 x 10*M ZnSO, Stock Solution 1 mL of 1.0 M Cuso, 1 mL of 1.0x 102 M Cuso, 1 mL of 1.0 M Znso, 1 mL of 1.0 × 10-2 M ZnSO4 5. Show the calculations for determining the volume of DI water to add to the given stock solution for the preparation of 1.0 x 10-2 M CuSO, and 1.0 x 10-4 M CuSO, solutions.Explanation / Answer
Desired solution 1.0 X 10-2 M CuSO4 from stock 1ml 1.0 M CuSO4 solution
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 1.0 M
V1 = initial volume = 1 ml
C2 = final concentration = 1.0 X 10-2 M
V2 = final volume = ?
C1V1 =C2V2 Thus,
V2 = C1V1 /C2
Substitute value in above equation
V2 = 1.0 X 1 / 1.0 X 10-2 = 100 ml
final volume of solution is = 100 ml
DI water to be add = final volume - initial volume = 100 - 1 = 99 ml
DI water to be add = 99 ml
Desired solution 1.0 X 10-4 M CuSO4 from stock 1ml 1.0 X 10-2 M CuSO4 solution
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 1.0 X 10-2M
V1 = initial volume = 1 ml
C2 = final concentration = 1.0 X 10-4 M
V2 = final volume = ?
C1V1 =C2V2 Thus,
V2 = C1V1 /C2
Substitute value in above equation
V2 = 1.0 X 10-2 X 1 / 1.0 X 10-4 = 100 ml
final volume of solution is = 100 ml
DI water to be add = final volume - initial volume = 100 - 1 = 99 ml
DI water to be add = 99 ml
Desired solution 1.0 X 10-2 M ZnSO4 from stock 1ml 1.0 M ZnSO4 solution
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 1.0 M
V1 = initial volume = 1 ml
C2 = final concentration = 1.0 X 10-2 M
V2 = final volume = ?
C1V1 =C2V2 Thus,
V2 = C1V1 /C2
Substitute value in above equation
V2 = 1.0 X 1 / 1.0 X 10-2 = 100 ml
final volume of solution is = 100 ml
DI water to be add = final volume - initial volume = 100 - 1 = 99 ml
DI water to be add = 99 ml
Desired solution 1.0 X 10-4 M ZnSO4 from stock 1ml 1.0 X 10-2 M ZnSO4 solution
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 1.0 X 10-2M
V1 = initial volume = 1 ml
C2 = final concentration = 1.0 X 10-4 M
V2 = final volume = ?
C1V1 =C2V2 Thus,
V2 = C1V1 /C2
Substitute value in above equation
V2 = 1.0 X 10-2 X 1 / 1.0 X 10-4 = 100 ml
final volume of solution is = 100 ml
DI water to be add = final volume - initial volume = 100 - 1 = 99 ml
DI water to be add = 99 ml
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