DILUTING MOLAR

SOLUTIONS

Diluting Solutions

Concentrated solutions have a relatively high molarity.

Dilute solutions have a relatively low molarity.

It is often faster to prepare a number of standard solutions by diluting a more

concentrated solution.

The following equation can be used to solve dilution problems – when water is

added or removed from a solution.

M1V1 = M2V2

M1= the initial molarity M2 = the final molarity

V1 = the initial volume V2 = the final volume

The Dilution Formula – How it Works

If we have 1 L of 3 M HCl, what is the new concentration if we dilute acid to 6 L (add 5 L of water to the solution)?

M1v1 = 3 mol

V1 = 1 LM1 = 3 M

v2 = 6 LM2 = 0.5 M

M2v2 = 3 mol

1 L x 3 mol = 3 mol HCl

1 L

3 mol HCl6 L

3 mol/1L

1. 25.0 mL of 0.10 M solution is diluted by adding 75.0 mL of water. Calculate

the molarity of the new solution.

M1V1 = M2V2

Initial Solution

M1 = 0.10 M M2 = ?

V1 = 25.0 mL V2 = 100.0 mL

25.0 mL + 75.0 mL

(0.10 M)(25.0 mL) = M2(100.0 mL)

M2 = 0.025 M

2. What volume of 0.250 M CoCl2 solution can be diluted to 100.0 mL in order to

make a 0.0150 M solution?

Initial Solution

M1V1 = M2V2

M1= 0.250 M M2 = 0.0150 M

V1 = ? V2 = 100.0 mL

(0.250 M) V1 = (0.0150 M)(100.0 mL)

V1 = 6.00 mL

3. 50.0 mL of 0.500 M CuSO4 solution is diluted to 250.0 mL. Calculate the new

concentration and the number of grams CuSO4 in the new solution.

M1 V1 = M2V2

M1= 0.500 M M2 = ?

V1 = 50.0 mL V2 = 250.0 mL

(0.500 M) (50.0 mL) = M2 (250.0 mL)

M2 = 0.100 M

0.250 L x 0.100 mol

1 L

x 159.6 g

1 mol

= 3.99 g

4. 1.53 g of NaCl is dissolved in 100.0 mL of water.

Calculate the molarity. The solution is concentrated by

evaporating off 60.0 mL of water. Calculate the new

molarity.

1.53 g 58.5 g

1 mol

x

0.1000 L

M1V1 = M2V2

(0.262 M)(100.0 mL) = M2(40.0 mL)

M2 = 0.654 M

100.0 mL – 60.0 mL

= 0.0262 mol NaCl

0.0262 mol NaCl = 0.262 M NaCl

Mixing Two Solutions

1.0 L of a 3.0 M HCl solution is added to 0.50 L

of a 2.0 M HCl solution. What is the final

concentration of HCl?

1.0 L 1 L

3.0 mol

x = 3.0 mol HCl

0.50 L 1 L

2.0 mol

x = 1.0 mol HCl

= 4.0 mol HCl

1.5 L

= 2.67 M HCl1.0 L + 0.50 L

TRY A FEW…

1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?

2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?

3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?

4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl?

5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl?

6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?

7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?

31.25 mL

1.2 M

0.4 L

2.6 M

0.38 M

0.3 M

16 L