Question

A standard curve was to be prepared from a stock of bromophenol blue. The stock concentration was 5 x 104 M. The standard concentrations required are listed in the table below, with the total volume of each to be 20 mL. 6. Table 1.2 Table for working out the volume of stock and water to be added for to make up the final concentrations required for Question 6 Vol. of stock concentration (mL) Vol. of water (mL) Final concentration (M) 0 5 x 104 3 x 10 1 x 104 5 x 10

Answer 1

You can prepare all the solutions by using the following equation, V1.N1 = V2.N2, where V1 and N1 are volume and concentration of the stock solution V2 and N2 are volume and concentration of the final solution In each case final volume = 20 ml and final concentrations are given in table, ie., the V2 and N2 are known. Again, we know the concentration of stock solution (N1) 5 X 104 M. We need to find out the volume that the stock solution required (V1) to prepare 20 ml of the given (in table) final concentration. Lets solve this by preparing 3 X 104 M solution. Here, V2 20 ml, N2 3 X 104 M, N1-5 X 104 M, V1-? V1.N1 V2.N2 Or, V1 V2.N2/N1 - 20 X 3 X 104/(5 X 104) 12 ml So, 12 ml of stock solution should be diluted upto 20 ml with water, ie., 12 ml stock solution and 8 ml water (net Similarly, each value can be calculated. The calculated values are given in table. 12+820 ml. 6. Astandard curve was to be prepared from a stock of bromophenol blue. The stock concentration was 5 x 104 M. The standard concentrations required are listed in the able below, with the total volume of each to be 20 mL Table 1.2 Table for working out the volume of stock and water to be added for to make up the final concentrations required for Question 6. Vol. of stock Vol. of water (mL) Final concentration (M) concentration (mL) 0 5 x 104 3 x 10 1 x 10 5 x 10 20 20 12 04 02 08 16 18