Five standard solutions of HBr are prepared by serial dilution in which, at each stage, 10.00...
Five standard solutions of HBr are prepared by serial dilution in which, at each stage, 10.00 mL are diluted to 160.00 mL. Given that the concentration of the most dilute solution is 2.62 × 10−6 M, determine the concentration of the original HBr stock solution.
Homework Answers
Given: Concentration of most dilute solution = 2.62 x 10-6 M
Volume of final solution
We will use the formula M1V1 = M2V2, to calculate the unknown concentration at each level
(Total 4 dilutions have been carried out and concentration after 4th dilution is given.)
(1) M1 = 2.62 x 10-6 M, V1 =160, V2 = 10 ml, M2 = unknown concentration
2.62 x 10-6 x 160 = M2 x 10
M2= (2.62 x 10-6 x 160) / 10 = 41.92 x 10-6 M = 0.419 x 10-4 M (This is concentration after third dilution)
(2) Again, using the same formula M2V2 = M3V3
M3= (0.419 x 10-4 x 160) / 10 = 6.704 x 10-4 M (Concentration after second dilution)
(3) Again, using the formula M3V3 = M4V4
M4 = 6.704 x 10-4 x 160 / 10 = 1.0726 x 10-2 M (Concentration after first dilution)
(4) Using last time, M4V4 = M5V5
M5= 1.0726 x 10-2 x 160 / 10 = 0.172 M
Concentration of original solution = 0.172 M
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