Everything Science Grade 11

CHAPTER 8. TYPES OFREACTIONS 8.1

The molarity of the NaOH solution is 0.12 mol.dm^3

or 0.12 M

Example 2: Titration calculation II

QUESTION

4.9 g of sulphuric acid is dissolved in water andthe final solution has avolume of 220 cm^3.

Using titration, it was found that 20 cm^3 of this solution was ableto completely neutralise 10

cm^3 of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in

mol.dm−^3.

SOLUTION

Step 1 : Write a balanced equation for the titration reaction.

H 2 SO 4 + 2NaOH→ Na 2 SO 4 + 2H 2 O

Step 2 : Calculate the molarityof the sulphuric acid solution.

M =

n

V

V = 220 cm^3 = 0. 22 dm^3

n =

m

M

=

4. 9 g

98 g· mol−^1

= 0. 05 mol

Therefore,

M =

0. 05

0. 22

= 0. 23 mol· dm−^3

Step 3 : Calculate the moles of sulphuric acid that were used in the neutralisation

reaction.

Remember that only 20cm^3 of the sulphuric acid solution is used.

M =nV, therefore n = M× V

n = 0. 23 × 0 .02 = 0. 0046 mol

Step 4 : Calculate the number of moles of sodium hydroxide that were neutralised.

According to the balanced chemical equation, the mole ratio of H 2 SO 4 to NaOH

is 1:2. Therefore, the number of moles of NaOHthat are neutralised is 0. 0046 ×

2 = 0.0092 mol.

Step 5 : Calculate the concentration of the sodium hydroxide solution.

M =

n

V

=

0. 0092

0. 01

= 0. 92 M