Preparing Solutions and Dilutions

Jan 2012
1
0
I am trying to figure out how to make the correct solutions for my chemistry experiment.

"To prepare your solutions, you will need to weight out quinine sulfate and dilute it in an aqueous solution of .05M H2SO4. Calculate the volume of H2SO4 (12mol/L) that you will need to prepare 250mL of a .05mol/L H2SO4 solution. We recommend that you prepare 50mL of a quinine stock solution that is 25 times more concentrated than your lowest solution concentration. Given the formula weight for quinine sulfate is 782.96g/mol calculate how many grams of quinine sulfate you will need. Calculate the volume of the stock solution you will put into these flasks."

The stock solution that I want to make that would be 25 more concentrated than the lowest concentration is .005moles/L

To prepare the 250mL of .05mol/L H2SO4 I did the following calculation:

(12mol/L)*x=(.05mol/L)*(250ml)
x= 1.04166mL of (12mol/L) H2SO4 to make (.05mol/L) H2SO4

My question is what do I add to the 1.04166mL of 12mol/L H2SO4 to dilute the solution to 250ml? So if I am using 1.04166mL of the 12mol/L H2SO4 what is the rest of the 249mL for the dilution?

My second question is if I want my stock solution of quinine sulfate to be .005mole/L does this mean that I will use (.005mol/L)*782.96(g/mol)= 3.9148grams of quinine sulfate to make this solution, as well as (.005mol/L)*(50mL)=(.05mol/L)*x where x is 5mL of .05mol/L H2SO4.

Therefore I would put 3.9148grams of quinine sulfate and then add 5ml of the .05mol/L H2SO4 and then dilute it with something to make 50ml of .005mol/L solution?

Again what do I use for the other 45mL of solution that isnt the 5ml of .05mol/L H2SO4

Any help or thoughts are greatly appreciated! Thank you
 
May 2010
491
8
prepare solutions

You are asked to prepare two solutions
1. 250 cc of 0.05 m/l H2SO4
2. 50 cc of 1.25 m/l quinine sulfate
Is this correct?