Avolume of 24.72±0.06 ml of hno3 solution was required for complete reaction with 0.8359±0.0007 g of na2co3, (fm 105.988±0.001 g/mol ). find the molarity of the hno3 solution and its absolute uncertainty.

Molarity of HNO₃ = 0.6379 ± 0.0020 M

Explanation:

Hi!

The equation of the reaction is as follows:

Na₂CO₃ + 2HNO₃ → H₂CO₃ + 2NaNO₃

The number of moles of Na₂CO₃ present in the reaction is:

moles Na₂CO₃ = 0.8359 ± 0.0007 g /  105.988 ± 0.001 g/mol  

First, let´s convert the absolute uncertainties in relative ones.

0.8359 ± 0.0007 g = 0.8359 ± ( (0.0007 g/ 0.8359 g) * 100%)

=0.8359g ± 0.08%

105.988 ± 0.001 g = 105.988 ± ((0.001 g/mol / 105.988 g/mol) * 100%)

= 105.988 g/mol ± 9×10⁻⁴%

Then:

moles Na₂CO₃ = (0.8359g ± 0.08%) /  (105.988 g/mol ± 9×10⁻⁴%)

moles Na₂CO₃ = (0.8359 g  / 105.988 g/mol) ± (0.08% + 9×10⁻⁴%)

moles Na₂CO₃ = 7.887 × 10⁻³ mol ± 0.08%

From the equation, we know that 2 moles of HNO₃ react with 1 mol of Na₂CO₃. Then, the number of moles of HNO₃ present in the reaction is:

moles of HNO₃ = 2 * moles of Na₂CO₃ = 2 * 7.887 × 10⁻³ mol ± 0.08%

moles of HNO₃ = 0.01577 mol ± 0.08%

This number of moles was present in 24.72 ml ± 0.2%. Then, in 1 l there will be:

moles of HNO₃ in 1l = 0.01577 mol ± 0.08% * 1000 ml / 24.72 ml ± 0.2%

moles of HNO₃ in 1l = 0.6379 mol ± (0.08% + 0.2%)

moles of HNO₃ in 1l = 0.6379 mol ± 0.3%

Converting the relative uncertainty to absolute one:

0.3% *  0.6379 mol / 100% = 2 × 10⁻³ mol

Then, the molarity of HNO₃ will be:

Molarity of HNO₃ = 0.6379 ± 0.0020 M