Question

4.24

Equal volumes of 0.050 M Ba(OH)₂ and 0.040 M   HCl are allowed to react. Calculate the molarity of each of the ions present after the reaction.

Answer: 0.025  M Ba²⁺  ; 0.020 M   HCl^- ; [H⁺] = (approx) 0 ; 0.30 M OH^-

Answer 1

Reaction is :

Ba(OH)₂ + 2HCl BaCl₂ + 2H₂O

or in ionic form:

Ba²⁺ + 2OH^- + 2H⁺ + 2Cl^- Ba²⁺ + 2 Cl^- + 2H₂O

This reaction shown that Ba²⁺ and Cl^- remains same in the solution before and after the reaction. But H⁺ and OH^- reacts with each other to give water which is non-electrolyte. so after reaction both H⁺ and OH^- diminishes.

Let Volume of each Ba(OH)₂ and HCl be 1 L

Molarity of Ba(OH)₂ = 0.050 M

Molarity of HCl = 0.040 M

Before mixing:

Moles of Ba(OH)₂ = Molarity * Volume = 0.050 M* 1L = 0.050 moles

Moles of HCl = Molarity * Volume = 0.040 M * 1 L = 0.040 moles.

After mixing

Total volume = 2 L

From reaction:

1 mole of Ba(OH)₂ reacts with 2 mole of HCl

Thus,  0.040 moles.of HCl reacts with 0.040 moles/ 2 = 0.020 moles of  Ba(OH)₂

Hence, Moles of HCl remains = 0 moles

Moles of Ba²⁺ and Cl^- remains same before and after the reaction i.e. 0.050 moles and 0.040 moles of Ba²⁺ and Cl^- respectively.

So, Molarity of Ba²⁺ = Moles/volume = 0.050 moles/ 2L = 0.025M

Molarity of Cl^-= Moles/volume = 0.040 moles/ 2L = 0.020 M

Since HCl reacts fully, SO, moles of H⁺ in solution = 0

Hence, Molarity of H⁺ = 0

Moles of OH^- in 0.050 moles of Ba(OH)₂ = 2* 0.050 = 0.100 moles

Moles of OH^- reacts = 0.040 moles

Thus, Moles of OH^- remains unreacted = 0.100 moles - 0.040 moles = 0.060 moles

Molarity of OH^- = 0.060 moles/ 2L= 0.030 M