Question

5.00mL of the BTB stock solution will be used in all 5 of the solutions to be studied. Each of the five solutions are diluted to 50.00 mL total. What is the new concentration of BTB in the solutions? (BTB stock concentration 2.5 x 10^-4 M)

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Solution: BTB KH2PO4 K2HPO4 DI water

1. 5.00 mL. 0.00 mL 10.00 mL 35.00 mL

2. 5.00 mL 5.00 mL    10.00 mL 30.00 mL

3 5.00 mL 10.00 mL   10.00 mL 25.00 mL

4 5.00 mL 15.00 mL   10.00 mL 20.00 mL

5 5.00 mL 10.00 mL 0.00 mL    35.00 mL

Answer 1

Ans. Final volume made in each case = 50.0 mL [see figure]

            Initial [BTB], C1 = 2.50 x 10^-4 M

            Initial volume of BTB stock solution taken, V1 = 5.0 mL

Now, using

            C1V1 (Stock BTB solution) = C2V2 (final solution)

            Or, 2.50 x 10^-4 M x 5.0 mL = C2 x 50.0 mL

            Or, C2 = (2.50 x 10^-4 M x 5.0 mL) / 50.0 mL

            Hence, C2 = 2.500 x 10^-5 M

Hence, [BTB] in final aliquot = 2.500 x 10^-5 M