A student does a titration using sodium carbonate solution and nitric acid.
The equation for the reaction is:
Na2CO3 + 2 HNO3 ->>> 2 NaNO3 + CO2 + H2O
25.0 cm³ of 0.124 mol/dm³ sodium carbonate solution is neutralised by
23.6 cm³ of nitric acid.
Calculate the concentration of the nitric acid.
Give your answer to 3 significant figures.
You should calculate:
- the number of moles of sodium carbonate in 25.0 cm³ of the solution
- the number of moles of nitric acid in 23.6 cm³ of the nitric acid
the concentration of the nitric acid in mol/dm³.
[5 marks]

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pianolover69plays pianolover69plays · Answer 1 · 2024-04-09T19:16:31+02:00

To solve this problem, we'll follow these steps:

1. Calculate the number of moles of sodium carbonate in 25.0 cm³ of the solution.

2. Use the stoichiometry of the reaction to determine the number of moles of nitric acid neutralized by the sodium carbonate.

3. Calculate the concentration of the nitric acid.

Given:

- Volume of sodium carbonate solution (V1) = 25.0 cm³

- Concentration of sodium carbonate (C1) = 0.124 mol/dm³

- Volume of nitric acid (V2) = 23.6 cm³

### Step 1: Calculate the number of moles of sodium carbonate:

Number of moles (n1) = concentration (C1) * volume (V1)

= 0.124 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³)

≈ 0.00310 mol

### Step 2: Use the stoichiometry of the reaction to determine the number of moles of nitric acid:

From the balanced equation:

1 mole of Na2CO3 reacts with 2 moles of HNO3.

So, the number of moles of HNO3 (n2) = 0.00310 mol * (2 mol HNO3 / 1 mol Na2CO3)

= 0.00620 mol

### Step 3: Calculate the concentration of the nitric acid:

Concentration of nitric acid (C2) = number of moles (n2) / volume (V2)

= 0.00620 mol / (23.6 cm³ / 1000 cm³/dm³)

≈ 0.262 mol/dm³

So, the concentration of the nitric acid is approximately 0.262 mol/dm³ (to 3 significant figures).