Question

What is the molar concentration of HCl in a solution made by reacting 178.5 mL of 2.461 M HCl with 7.140 g of Na2CO3?

Answer 1

Solution: Let us first calculate the amount of HCl present in the 178.5 mL of 2.46 M HCl solution. The molecular weight of HCl is 36.

As

36 g HCl in 1000 mL = 1M

2.46 x 36 g HCl in 1000 mL = 2.46 M

88.56 g HCl in 1000 mL = 2.46 M

Now calculate the amount of HCl in 178.5 mL solution

As, 1000 mL = 88.56 g HCl

For 178.5 mL = (178.5 x 88.56) / 1000 g of HCl

= 15.80 g

Therefore 178.5 mL of 2.46 M HCl solution contains 15.80 g of HCl. The number of moles of HCl can be calculated using following formula,

No. of moles of HCl = Weight of HCl (g) / Molecular Weight of HCl

= 15.80 (g) / 36

= 0.39 moles of HCl

Similarly calculate number of moles of Na₂CO₃. The weight and molecular weight of Na₂CO₃ are 7.140 g and 106, respectively.

No. of moles of Na₂CO₃ = Weight of Na₂CO₃ (g) / Molecular Weight of Na₂CO₃

= 7.140 g / 106

= 0.0673 moles

The reaction of Na₂CO₃ with HCl is shown below

It is clear from the reaction that 1 mole of Na₂CO₃ will consume 2 moles of HCl. Let us calculate the number of moles of HCl required to neutralize 0.0673 moles of Na₂CO₃

1 mole of Na₂CO₃ = 2 moles of HCl

Therefore,

0.0673 moles of Na₂CO₃ = 0.0673 x 2

= 0.1347 moles of HCl

Thus, out of total 0.39 moles of HCl, 0.137 moles will be consumed to neutralize 0.0673 moles of Na₂CO₃. The remaining amount of HCl will be,

0.39 moles - 0.1347 moles = 0.225 moles

So, 178.5 mL solution will have 0.225 moles of HCl after treatment with 7.140 g of Na₂CO₃.

Now calculate the number of moles of HCl in 1000 mL.

As, 178.5 mL = 0.225 moles

1000 mL = 1.43 moles

Therefore, the molar concentration of HCl will be 1.43 M after reaction with 7.140 g of Na₂CO₃.