Chapter 15, Problem 36A

Interpretation Introduction

Interpretation: The volume of the given solution is to be calculated to form the solution.

Concept Introduction:

The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions.

Answer to Problem 36A

The required volume is $7.\text{5mL}$ .

Explanation of Solution

Given information:

The volume of barium nitrate is, $V_{{\text{Ba}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}=12.\text{5mL}$

The molarities of the barium sulphate and sodium sulphate are,

  $\begin{array}{l}{M}_{{\text{Ba}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}=0.1\text{5M}\\ M_{{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}=0.\text{25M}\end{array}$

Calculation:

The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

By substituting the given values in the above formula and then the result is,

  $\begin{array}{c}{M}_{{\text{Ba}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{V}_{{\text{Ba}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}=M_{{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}{V}_{{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ V_{{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}=\frac{M_{{\text{Ba}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{V}_{{\text{Ba}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}}{M_{{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}}\\ =\frac{\text{0}\text{.15M×12}\text{.5mL}}{\text{0}\text{.25M}}\\ \text{=}\text{7}\text{.5mL}\end{array}$

Conclusion

The required volume of $\text{0}\text{.25M}$ solution is $\text{7}\text{.5mL}$ .