p. 31 Molarity By Dilution (mixing a concentrated solvent
with water)
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)
C1
How much concentrated 18 M sulphuric acid is needed to prepare
250.0 mL of a 6.0 M solution?
V2
C2
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 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 acid
V1= 50.0 mL-amount of concentrated acid
C2= 4.0 M-diluted acid
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 acid
V1= 100.0 mL-amount of concentrated acid
C2= 1.5 M-diluted acid
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