Molarity by Dilution

p. 31 Molarity By Dilution (mixing with water

a concentrated solvent)

Acids are usually acquired from chemical supply houses in concentrated form. These acids are diluted to the desired concentration by adding water. Since moles of acid before dilution = moles of acid after dilution, and moles of acid : CV then,

C1 X V1 = C2 X V2. Solve the following problems.

1) 83 mL 2) 17 mL 3) 130 mL 4)1.0 X102 mL 5) 1100 mL

1) How much concentrated 18 M sulphuric acid is needed to prepare
250.0 mL of a 6.0 M solution?
C1V1 =C2V2 ® V1 = C2V2
C1
C1=18 M –concentrated
V1= how much volume?
C2=6.0 M -diluted
V2=250.0 mL –total volume
V1 = 6.0 M X 250.0 mL = 83 mL or 0.083 L
18 M
2) How much concentrated 12 M hydrochloric acid is needed to prepare
100.0 mL of a 2.0 M solution?
V1 = C2V2
C1
C1=12 M -concentrated
C2=2.0 M-diluted
V2=100.0 mL-total volume
V1 = 2.0 M X 100.0 mL = 17 mL or 0.017 L
12 M
3) To what volume should 25 mL of 15 M nitric acid be diluted to prepare
a 3.0 M solution?
V2 = C1V1
C2
C1=15 M-concentrated
C2=3.0 M-diluted
V1=25 mL
V2 = 15 M X 25.0 mL = 125 mL = 130 mL = 0.13 L
3.0 M
*T*4) To how much water should 50.0 mL of 12 M hydrochloric acid be
added to produce a 4.0 M solution?
C1= 12 M-concentrated
V1= 50.0 mL-amount of concentrated
C2= 4.0 M-diluted
V2 = C1V1
C2
V2 = 12 M X 50.0 mL = 150 mL ¬ Total Volume
4.0 M
V2 - V1
V(water) = 150 mL - 50.0 mL = 100 mL = 1.0 X 102 mL
*T*5) To how much water should 100.0 mL of 18 M sulphuric acid be
added to prepare a 1.5 M solution?
C1= 18 M-concentrated
V1= 100.0 mL-amount of concentrated
C2= 1.5 M-diluted
V2 = C1V1
C2
V2 = 18 M X 100.0 mL = 1200 mL ¬ Total Volume
1.5 M
V2 - V1
V(water) = 1200 mL - 100.0 mL = 1100 mL = 1.1 X 103 mL = 1.1 L