GCSE / O Levels Chemistry

4.0 g of sodium hydroxide, NaOH, is dissolved in 250 cm³ of water in a graduated flask.
A 25 cm³ sample of this solution is titrated with 0.50 mol/dm³ hydrochloric acid.
Which volume of hydrochloric acid is required to exactly neutralise the alkali?
A 10 cm³
B 20 cm³
C 40 cm³
D 200 cm³

Choice B is correct because:

The balanced equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is:

NaOH + HCl -> NaCl + H2O

From the balanced equation, it can be seen that 1 mole of NaOH reacts with 1 mole of HCl.

Given that the concentration of HCl is 0.50 mol/dm³ and the question uses a 25 cm³ sample of the NaOH solution, determine the number of moles of NaOH in the sample.

Step 1: Convert the volume of the NaOH solution to dm³: 

Volume of NaOH solution = 25 cm³ 

Volume of NaOH solution = 25/1000 dm³ 

Volume of NaOH solution = 0.025 dm³

Step 2: Calculate the number of moles of NaOH: 

Number of moles of NaOH = mass / molar mass Mass of NaOH 

Molar mass of NaOH = 23 g/mol + 16 g/mol + 1 g/mol = 40.0 g/mol 

Number of moles of NaOH = 4.0 g / 40.0 g/mol 

Number of moles of NaOH = 0.1 mol

Step 3: Determine the volume of HCl required using the mole ratio: 

From the balanced equation, 1 mole of NaOH reacts with 1 mole of HCl. 

Therefore, the number of moles of HCl required is also 0.1 mol.

Now, calculate the volume of HCl:

 Volume of HCl = (Number of moles of HCl) / (Concentration of HCl) 

Volume of HCl = 0.1 mol / 0.50 mol/dm³ 

Volume of HCl = 0.20 dm³

Finally, convert the volume to cm³:

 Volume of HCl = 0.20 dm³ x 1000 cm³/dm³ 

Volume of HCl = 20 cm³

Choices A, C and D are incorrect because they do not state the correct volume of HCl required to neutralize the alkali.

Experimental Chemistry | 2022 - O/N - 5070/12 | Q15