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?

Given:

Molarity of HCl, = 0.1 M

Mass of Mixture = 1g

To find: Volume of HCl to react completely with mixture

Formula:

Molarity, M_o

Solution:

Calculation of Amount of each component in mixture:

⇒ Let the amount of Na₂CO₃ be X g

⇒ And Let amount of NaHCO₃ be [1-X] g

⇒ Molecular Weight of Na₂CO₃ = [23 × 2] + [12] + [3 × 16]

= 106 g

⇒ Molecular Weight of NaHCO₃ = [23] + [1] + [12] + [3 × 16]

= 84 g

⇒ Number of moles of NaHCO₃

⇒ Number of moles of Na₂CO₃

Now it is given in the question that the mixture is equimolar, so

⇒ Number moles of Na₂CO₃ = Number of moles of NaHCO₃

106 – 106X = 84X

190X = 106

X = 0.5579

≈ 0.558

Amount of Na₂CO₃ = 0.558 g

⇒ Amount of NaHCO₃ = 1- 0.558 g = 0.442 g

⇒ Number moles of Na₂CO₃ = [0.558/106]

= 0.00526 moles

⇒ Number of moles of NaHCO₃ = 0.00526 moles …. [Mixture is equimolar]

Calculation of volume of HCl required:

⇒ In the question it is given that HCl reacts with the mixture.

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

NaHCO₃ + HCl → NaCl + H₂O + CO₂

From the above chemical equations it can be found that 2 moles of HCl is required to react with one mole of Na₂CO₃ and 1 moles of HCl is required to react with one mole of NaHCO₃.

So total number of moles of HCl required to completely react with mixture is as follows:

⇒ Total Number of Moles of HCl = [2 × 0.00526] + 0.00526

= 0.01578 moles

From Molarity Formula we have,

Molarity, M_o

⇒

⇒ Volume in litres, V = [0.01578/0.1]

= 0.1578 L

⇒ Volume in ml, V = 0.1578 × 1000

= 157.8 ml

Therefore the volume of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both is 157.8 ml.