Question

Constant-boiling HCl can be used as a primary standard for acid-base titrations. A 50.0 mL sample of constant-boiling HCl with a concentration of 0.1166 M was collected and titrated to an end point with 37.17 mL of Ba(OH)2 solution.

 

What is the molarity of the Ba(OH)2 solution?

Answer 1

Concepts and reason

Molarity (M) is calculated as the number of moles (n) divided by the volume (V).

$M = \frac{n}{V}$

1 mole consists of $6.022 \times {10^{23}}$ atoms. This is called Avogadro’s Number.

Fundamentals

The reaction between hydrochloric acid and Barium hydroxide is as follows:

${\rm{2HCl + Ba}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}\longrightarrow{{}}{\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Calculate the number of moles of HCl $\left( {{n_{{\rm{HCl}}}}} \right)$ from the volume of HCl (V) and concentration of HCl (c)as follows:

${n_{{\rm{HCl}}}} = Vc$

Substitute V as 50 mL and c as 0.1166 M in the above expression as follows:

$\begin{array}{c}\\{n_{{\rm{HCl}}}} = \left( {50{\rm{ mL}}} \right)\left( {\frac{{1{\rm{ L}}}}{{1000{\rm{ mL}}}}} \right)\left( {0.1166{\rm{ M}}} \right)\\\\ = 0.00583{\rm{ mol}}\\\end{array}$

Calculate the number of moles of Barium hydroxide $\left( {{n_{{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_2}}}} \right)$ as follows:

${n_{{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}} = \frac{1}{2}{n_{{\rm{HCl}}}}$

Substitute ${n_{{\rm{HCl}}}}$ as 0.00583 mol in the above expression as follows:

$\begin{array}{c}\\{n_{{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}} = \frac{1}{2}\left( {0.00583} \right){\rm{ mol}}\\\\ = 0.002915{\rm{ mol}}\\\end{array}$

Molarity of Barium hydroxide is calculated as follows:

$M = \frac{{{n_{{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_2}}}}}{V}$

Substitute ${n_{{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_2}}}$ as 0.002915 mol and V as 37.17 mL in the above expression as follows:

$\begin{array}{c}\\M = \frac{{0.002915{\rm{ mol}}}}{{37.17{\rm{ mL}}\left( {\frac{{1{\rm{ L}}}}{{{\rm{1000 mL}}}}} \right)}}\\\\ = \frac{{0.002915{\rm{ mol}}}}{{0.03717{\rm{ L}}}}\\\\ = 0.07842{\rm{ M}}\\\end{array}$

Ans:

The molarity of ${\rm{Ba}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}$ solution is 0.07842 M.