The mole concept and solutions
Francine Taylor-Campbell, Contributor

Reverend Al Miller and Christopher 'Johnny' Daley (M.C.) join musicians on-stage, at Peace Day celebrations, at Vauxhall High School, on March 6. - Contributed

Main points

• Concentration is expressed as the amount of solute in a given volume of solution.
  
  
• Concentration is expressed in units of moles of solute in 1dm³ of solution (mol/dm³).
  
  
• Mass concentration is the mass of the solute in 1dm³ of solution (g/dm 3 ).

A standard solution is a solution, the concentration of which, in mol/dm3 is known or can be calculated.

e.g. Calculate the mass of sodium carbonate needed to make up 500cm 3 of a 0.20M solution.

M (molarity) is the number of moles in 1dm³ , that is mol/dm³.

Thus, 0.2M = 0.2 mol in 1000cm³ (1dm³)

X mol == 500cm³

X = (500x0.2)/1000 = 0.1 mol

Mr of Na2 CO3 = 106g, thus the mass of 0.1 mol = 106x0.1 = 10.6g

10.6g of Na2 CO3 dissolved in 500cm3 has a concentration of 0.2M.

Alternative method

What mass of NaOH is needed to make up 250cm 3 of a 2M solution?

2M = 2 mol in 1000cm³ (1dm³ )
Mr of NaOH = 40g then 2 mol == 80g
80g = 1000cm³
Xg = 250cm³
X = (250*80)/1000 = 20g of NaOH

Concentration calculation

As stated earlier, concentration can be expressed in mol/dm3 and g/dm3.

e.g. What is the concentration of 20cm 3 of sulphuric acid containing 0.25 mol H2 SO4?

0.25 mol == 250 cm³
x mol == 1000cm³ (1dm³)
x = (0.25x1000)/250 = 1 mol Concentration = 1 mol/dm 3

Calculate the concentration in g/dm 3 of the same solution?

Since, 0.25 mol H2 SO4 is present in 250cm3
Mr of H2 SO4 = 98g mass of 0.25 mol = 98x0.25 = 24.5g
Thus, 24.5g === 250cm³
X g ==== 1000cm³
X = (1000x24.5)/250 = 98g Concentration = 98g/dm³

Alternative method

Since the concentration = 1 mol/dm³

Concentration in g/dm³ = 1 mol/dm³ x Mr = 1 mol/dm³ x 98 g/mol = 98g/dm³

Q1 . Find the concentration in g/dm³ and mol/dm³ of the following. Solutions

(i) 30 cm³ of a nitric acid solution containing 0.10mol HNO³

(ii) 200cm³ of a sodium hydroxide solution containing 40g of NaOH

(iii) 40cm³ of an ammonium nitrate solution containing 16g of NH4 NO3

(iv) 350 cm³ of a sulphuric acid solution containing 0.185 mol H2 SO4.

The real test of this topic comes in its application to titration problems.

Consider the following: 24 cm³ of H2SO4 of concentration 5.20g/dm³ was titrated against NaOH solution and required 25 cm³ of NaOH for complete neutralisation. Calculate the concentration of the NaOH solution.

First of all, a balanced equation is needed to determine the mole ratio in which the reactants combine

2NaOH (aq) + H2 SO4 (aq) = Na2SO4 (aq) + 2H2 O(l)

2 mol of NaOH react with 1 mol of H2 SO4. Ratio is 2:1

Concentration of H2 SO4 in mol/dm³ = (5.20g/dm³)/98g/mol = 0.053mol/dm³

Next, calculate the number of moles of H2 SO4 in 24cm3

0.053mol H2 SO4 == 1000cm3 (1dm³)

x mol == 24cm³

x == (24x0.053)/1000 = 0.00127 mol

Since NaOH reacts with H2 SO4 in the ratio 2:1

Then, the number of moles of NaOH that would have reacted = 0.00127 x 2= 0.00254mol

0.00254 mol NaOH is present in 25cm³

x mol === 1000cm³

x = (1000x0.00254)/25 = 0.102 mol

Concentration = 0.102 mol/dm³

Concentration in g/dm³ = 0.102 mol/dm³ x Mr = 0.102 x 40 = 4.07g/dm³

Francine Taylor-Campbell is an independent contributor.