Question

How many mL of 0.1 M HCl are required to react completely with 1 g mixture Of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Answer 1

Let the mixture contains x g of sodium carbonate and (1−x) g of sodium bicarbonate.

The molar masses of sodium carbonate and sodium bicarbonate are 106 g/mol and 84 g/mol respectively 

The number of moles of sodium carbonate and sodium bicarbonates are \(\frac x{106}\)​ and \(\frac {1 -x}{84}\)​ respectively.

Since, it is an equimolar mixture,

\(\frac x{106} = \frac{1-x}{84}\)

84x = 106 − 106x

190x = 106

x = 0.5579

Number of moles of sodium carbonate = \(\frac{0.5579}{106} = 0.005263\)

Number of moles of sodium hydrogen carbonate = \(\frac{1-0.5579}{84} = 0.005263\)

One mole of sodium carbonate will react with 2 moles of HCl  and 1 mole of sodium bicarbonate will react with 1 mole of HCl.

Total number of moles of HCl that will completely neutralize the mixture = 2 × 0.005263 + 0.005263 = 0.01578 moles

Volume of 0.1 M HCl required = \(\frac{0.01578}{0.1}\)

= 0.158L

= 158 mL.

Answer 2

Step1: To calculate the number of moles of the components in mixture suppose Na₂CO₃ present in the mixture = X g, 

NaHCO₃ present in the mixture = (1 - x) g 

Molar mass of Na₂CO₃ = 2 x 23 +12 + 3 x 16 = 106g mol^-1 

Molar mass of NaHCO₃ = 23 + 1 + 12 + 3 × 16 = 84g mol^-1 

∴ Moles of Na₂CO₃ in xg = \(\frac{x}{106}\)

Moles of NaHCO₃ in (1-x) g = \(\frac{1 - x}{84}\)

Step 2: To calculate the moles of HCl required. 

Na₂CO₃ + 2HCl → 2 NaCl + H₂O + CO₂ 

NaHCO₃ + HCl → NaCl + H₂O + CO₂ 

1 mole of Na₂CO₃ required HCl = 2 moles 

0.00526 mole of Na₂CO₃ requires HCl 

= 0.00526 × 2 moles = 0.01052 

1 mole of NaHCO₃ required HCl = 1 mole 

0.00526 mole of NaHCO₃ required HCl = 0.00526 mole 

∴ Total HCl required = 0.01052 + 0.00526 =0.01578 moles 

Step 3: To calculate volume of 0.1M HCl 

0.1 mole of 0.1 M HCl are present in 1000 mL of HCl 

0.01578 mole of 0.1 M HCl will be present in HCl = \(\frac{1000}{0.1}\) x 0.01578 = 157.8 ml.