Question

25mL of Barium Hydroxide solution on titration with a 0.1 molar solution of Hydrochloric acid gave a tire value of 35mL. The molarity of Barium Hydroxide solution is?

Answer 1

This question is based upon the neutralization reaction of the base Barium Hydroxide and Acid- Hydrochloric acid. Both the acid and base given are strong in nature, hence will lead to complete neutralization reaction and formation of the salt Barium Chloride.

Complete answer:
The balanced chemical equation for the reaction that occurs can be given as:
$Ba{(OH)_2} + 2HCl \to BaC{l_2} + {H_2}O$
One mole of Barium Hydroxide will require two moles of Hydrochloric acid. In analytical chemistry for titrations the concentrations are commonly expressed in terms of normality. The relationship between Normality and Volume can be given as: $N \propto \dfrac{1}{V}$
Normality is inversely proportional to volume.
Let ${N_1}\& {V_1}$ be the normality and volume of the acid solution and ${N_2}\& {V_2}$ be the normality and volume of the basic solution. From the above solution we can say that:
${N_1} = \dfrac{k}{{{V_1}}}$ -- (1)
${N_2} = \dfrac{k}{{{V_2}}}$ -- (2)
From (1) and (2) we get that ${N_1}{V_1} = {N_2}{V_2}$ -- (3)
The information given to us is: ${V_{Ba{{(OH)}_2}}} = 25mL,{V_{HCl}} = 35mL,{M_{HCl}} = {N_{Hcl}} = 0.1N$
Substituting the values to get the normality of Barium Hydroxide: $0.1 \times 35 = {N_{Ba{{(OH)}_2}}} \times 25$
${N_{Ba{{(OH)}_2}}} = \dfrac{{0.1 \times 35}}{{25}} = 0.14N$
Now we know that One mole of Barium Hydroxide will require two moles of Hydrochloric acid, hence the Molarity of Barium hydroxide will be half of the normality found; 0.14N.
The molarity of Barium hydroxide will be = $\dfrac{{0.14}}{2}M = 0.07M$
The correct answer is 0.07M.

Note:
The alternate way to solve this problem is by using the formula: $\dfrac{{{M_{HCl}} \times {V_{HCl}}}}{{{n_{HCl}}}} = \dfrac{{{M_{Ba{{(OH)}_2}}} \times {V_{Ba{{(OH)}_2}}}}}{{{n_{Ba{{(OH)}_2}}}}}$
One mole of Barium Hydroxide needs two moles of Hydrochloric Acid. No. of moles of Hydrochloric acid will be one and that of barium hydroxide will be two.
Substituting the values in the formula and finding the molarity of Barium Hydroxide;
$\dfrac{{25 \times {M_{Ba{{(OH)}_2}}}}}{1} = \dfrac{{35 \times 0.1}}{2}$
${M_{Ba{{(OH)}_2}}} = 0.07M$ .