Solutions

Mixing two solutions:

Resolution Method and Exercises

Mixing two Solutions

Main Ideas

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When two solutions (1 and 2) are mixed to yield a resulting solution (3), some magnitudes can be added: they are called additive magnitudes:

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m₁ + m₂ = m₃

n₁ + n₂ = n₃

V₁ + V₂ = V₃

Mixing two solutions: Exercise

Exercise 1

450 mL 40 g/L KOH solution is added to a 200 mL 1.2 M KOH solution. Determine the concentration of the resulting solution in g/L and mol/L.

1: initial solution

3: final solution

2: matter added

200 mL�1.2 M

450 mL�40 g/L

c?�g/L and M

Mixing two solutions: Exercise

Method

The process involves steps of the following kinds:

a) Determination of additive magnitudes (magnitudes that can be added: m, n, V)

b) Addition of magnitudes in order to know final values

c) Determination of final concentration

Resolution

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1. The volume of final solution is the sum of volumes of initial and added solutions

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2. The initial number of moles of solute is the concentration (in moles per liter) times its volume.

Resolution

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3. The mass of solute (KOH) of initial solution is the number of moles times the molar mass (the mass of one mole).

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4. The mass of solute (KOH) of added solution is the concentration (mass per liter) times the volume of solution).

Resolution

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5. The number of moles of solute (KOH) of added solution is the mass over the molar mass.

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6. The mass of solute (KOH) of the resulting solution is the sum of masses of solute of initial and added solutions.

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7. The number of moles of the final solution is the sum of nomber of moles of initial and added solutions.

Resolution

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8. The concentration in grams per liter of the final solution is the ratio between the mass of solute over the volume of solution.

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9. The concentration in moles per liter of the final solution is the ratio between the number of moles of solute over the volume of solution.

Mixing two solutions: Exercises

Exercise 2

Calculate the volume of water we need to add to an 0.4 L 8 M initial solution in order to get a 2.5 M solution.

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Exercise 3

A container has a 40 % HNO₃ solution. Its density is d=1.25 g/mL. Calculate the volume we need to take from this container in order to get 500 mL of a solution 0.3 M by dilution.

Atomic weights: N=14; H=1; O=16

Mixing two solutions: Exercises

Exercise 2

Calculate the volume of water we need to add to an 0.4 L 8 M initial solution in order to get a 2.5 M solution.

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m = g solute

n = mol solute

V = 0.4 L solution

c = g /L

c = 8 mol/L

c = %

1: initial solution

m = g solute

n = mol solute

V = L solution

c = g /L

c = 2.5 mol/L

c = %

3: final solution

2: matter added

n = 0 mol solute

V = L solution

c = 0 g /L

c = 0 mol/L

c = 0 %

+

=

water

1

m = 0 g solute

2

3

4

Mixing two solutions: Exercises

Exercise 2

Calculate the volume of water we need to add to an 0.4 L 8 M initial solution in order to get a 2.5 M solution.

​

Mixing two solutions: Exercises

Exercise 3

A container has a 40 % HNO3 solution. Its density is d=1.25 g/mL. Calculate the volume we need to take from this container in order to get 500 mL of a solution 0.3 M by dilution.

m = g solute

n = mol solute

V = L solution

c = g /L

c = mol/L

c = 40 %

1: initial solution

m = g solute

n = mol solute

V = 0.5 L solution

c = g /L

c = 0.3 mol/L

c = %

3: final solution

2: matter added

m = 0 g solute

n = 0 mol solute

V = L solution

+

=

1

2

3

4

5

Mixing two solutions: Exercises

Exercise 3. A container has a 40 % HNO3 solution. Its density is d=1.25 g/mL.

100 g

40 g

1 L

1250 g

density

1.25 g/mL

ratio

40 / 100

Mixing two solutions: Exercises

Exercise 3. A container has a 40 % HNO3 solution. Its density is d=1.25 g/mL.

mass of solute (g)

mass of solution (g)

volume of solution (L)

1 L

100 g

1250 g

0.08 L

40 g

500 g

y = mx

0.08 L

y = mx

density = 1250 g / L

mass (solute) = 0.4 mass (solution)

0.4 means 40%

Mixing two solutions: Exercises

Exercise 3

A container has a 40 % HNO3 solution. Its density is d=1.25 g/mL. Calculate the volume we need to take from this container in order to get 500 mL of a solution 0.3 M by dilution.