A stock solution of CuSO4•5H2O was prepared by dissolving 1.6503 g of copper (II) sulfate pentahydrate in sufficient water to make 250.00 mL of solution.
a) Calculate the molarity of the stock solution with respect to copper (II) sulfate.
b) If a 25.00 mL aliquot of the stock solution is introduced into a 500.00 mL volumetric flask, calculate the molarity of the new solution with respect to Cu 2+ ion.
c) Calculate the volume of distilled water that must be added to prepare 20.00 mL volumes of each of the following CuSO4 solutions:
(i) 0.0200 M
(ii) 0.0100 M
(iii) 0.0050 M
Here are the answers, I just don't know how to get them:
a) Molarity of stock solution: [CuSO4] = 0.026442 M
b) [Cu 2+] = 1.3221 x 10 -3
c)(i) 4.87 mL of distilled water must be added.
(ii) 12.43 mL of distilled water must be added.
(iii) 16.21 mL of distilled water must be added.
I got a), so there's no need to show me how to get that answer.
a) Calculate the molarity of the stock solution with respect to copper (II) sulfate.
b) If a 25.00 mL aliquot of the stock solution is introduced into a 500.00 mL volumetric flask, calculate the molarity of the new solution with respect to Cu 2+ ion.
c) Calculate the volume of distilled water that must be added to prepare 20.00 mL volumes of each of the following CuSO4 solutions:
(i) 0.0200 M
(ii) 0.0100 M
(iii) 0.0050 M
Here are the answers, I just don't know how to get them:
a) Molarity of stock solution: [CuSO4] = 0.026442 M
b) [Cu 2+] = 1.3221 x 10 -3
c)(i) 4.87 mL of distilled water must be added.
(ii) 12.43 mL of distilled water must be added.
(iii) 16.21 mL of distilled water must be added.
I got a), so there's no need to show me how to get that answer.
-
b) Use the dilution equation
V1C1 = V2 C2
where
V1 = initial volume = 25.00 ml
C1 = initial concentration = 0.026442 M
V2 = final volume = 500.00 ml
C2 = ? M
C2 = V1C1 / V2
= 25.00 ml x 0.026442 M / 500.00 ml
= 0.0013221 M
= 1.322x10^-3 M
c) Use the dilution equation again. Use it to determine the volume of stock solution needed to produce the 20.00ml of the required solution. Then subtract this from 20.00 ml to give the volume of water needed (we have to assume the volumes are addative for this question.
i)
V1 = ? ml
C1 = 0.026442 M
V2 = 20.00 ml
C2 = 0.0200 M
V1 = C2V2 / C1
= 0.0200 M x 20.00 ml / 0.026442 M
= 15.127 ml
volume water = 20.00 ml - 15.127 ml
= 4.873 ml
= 4.87 ml
repeat for the other 2...
It would have been nice if the question had specified that it wanted you to prepare the solutions in c) from the stock solution, however, it is not possibe to make any of these from the solution you prepared in part b) since it is much less concentrated, so I guess I'm just eing a bit pedantic.
V1C1 = V2 C2
where
V1 = initial volume = 25.00 ml
C1 = initial concentration = 0.026442 M
V2 = final volume = 500.00 ml
C2 = ? M
C2 = V1C1 / V2
= 25.00 ml x 0.026442 M / 500.00 ml
= 0.0013221 M
= 1.322x10^-3 M
c) Use the dilution equation again. Use it to determine the volume of stock solution needed to produce the 20.00ml of the required solution. Then subtract this from 20.00 ml to give the volume of water needed (we have to assume the volumes are addative for this question.
i)
V1 = ? ml
C1 = 0.026442 M
V2 = 20.00 ml
C2 = 0.0200 M
V1 = C2V2 / C1
= 0.0200 M x 20.00 ml / 0.026442 M
= 15.127 ml
volume water = 20.00 ml - 15.127 ml
= 4.873 ml
= 4.87 ml
repeat for the other 2...
It would have been nice if the question had specified that it wanted you to prepare the solutions in c) from the stock solution, however, it is not possibe to make any of these from the solution you prepared in part b) since it is much less concentrated, so I guess I'm just eing a bit pedantic.