Question

add 100 mg quinine sulfate dihydrate and 50 ML 1M H2S04 to a 1.000 L volumetric flask in and then dilute to volume with distilled water. The resulting is 82.9 mg/L. (Ppm) in quinine. prepare the following quinine Solutions from the 82.9 mg/L standard. Used 0.05 M H2S04 to make the dilutions. make about 100 ML of the 8.3 mg/L solution and 50 of the others under.

8.3 mg/L

0.83 mg/L

8.3 X 10^-2 mg/L

8.3 X 10^-3 mg/L

Can someone explain to me how I would make these dilutions with the answer for each thanks!

Answer 1

Given stock solution = 82.9 mg/L, that means 82.9 ppm

Required solution 1) 8.3 ppm in 100 ml

We can use the formula M1V1 = M2V2

M1 = Concentration of stock solution = 82.9 ppm

V1= Volume of stock solution required = X

M2= Concentration of required solution = 8.3 ppm

V2 = Volume of required solution = 100 ml

V1 = M2V2 / M1 = ( 8.3 x 100 ) / 82.9 = 10.0 ml

Preparation: Take 10 ml of stock (82.9 ppm) solution and dilute it to 100 ml with 0.05M H2SO4 then you will get 8.3 ppm solution.

Similarly if calculated the required volume for other solutions we will get

2) For 0.83 ppm in 50 ml : Required volume of stock solution to dilute = ( 0.83 x 50) / 82.9 = 0.5 ml

3) For 8.3 X 10^-2 ppm in 50 ml : Required volume of stock solution to dilute

    = ( 8.3 X 10^-2x 50) / 82.9 = 0.05 ml

4) For 8.3 X 10^-3 ppm in 50 ml : Required volume of stock solution to dilute

    = ( 8.3 X 10^-3x 50) / 82.9 = 0.005 ml