Question

8. Describe briefly how you would prepare, with a minimum number of dilutions, a 1x 10 Msolution of NaOH given the following: a 0.08 M stock solution of NaOH; 2, 5, 10 mL pipettes; 10, 20, 50, 100 mL volumetric flasks. (Show your working)

Answer 1

Molarity of stock solution, M1 = 0.08 M

Let, Volume of stock used be  V1

Molarity of final solution, M2 = 0.001 M

Let, Volume of the final solution be  V2

moles of NaOH taken, n = M1 * V1

      = 0.08 * V1

moles of NaOH in final solution = M2 * V2

= 0.001 * V2

moles of NaOH in final solution = moles of NaOH taken from stock solution

0.001 * V2 = 0.08 * V1

0.001 = 0.08 * (V1 / V2)

Consider V1/V2 as the dilution ratio that can done with 2 or more dilutions : (V1/V)* (V/V2) so on.

also, V1/V2 = 0.001/0.08 = 1/80

Dilution ratio for this procedure is (volume of pippete / volume of flask)

Dilution ratio available to us:

	10 mL	20 mL	50 mL	100 mL
2 mL	2/10 = 1/5	2/20 = 1/10	2/50 = 1/25	2/100 = 1/50
5 mL	5/10 = 1/2	5/20 = 1/4	5/50 = 1/10	5/100 = 1/20
10 mL	10/10 = 1	10/20 = 1/2	10/50 = 1/5	10/100 = 1/10

For minimum dilutions, we can use the 2 dilution from the table above that are bolded i.e. 1/4 and 1/20

V1/V2 = 1/4 * 1/20 = 1/80 (our required total dilution)

Ans: For minimum dilution, we take 5mL stock solution and pour it in 20 mL flask and dilute it, then take 5mL of the resulting solution and pour it in 100 mL flask and dilute it to 100 mL.