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 |
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