Chapter 15, Problem 32A

Interpretation Introduction

Interpretation: The initial volume of the given solutions is to be determined.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• 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 32A

The respective initial volumes are $\text{55}\text{.78mL,}\text{42}\text{.45mL,}\text{38}\text{.57mL}\text{and}\text{45}\text{.3mL}$ .

Explanation of Solution

Given information:

The final volume is, $V_2=\text{225 mL}$ and final molarity is, $M_2=3.\text{0M}$

The individual initial molarities for each solutionis,

  $\begin{array}{l}{M}_1\left(\text{HCl}\right)\text{=}\text{12}\text{.1M}\\ M_1\left({\text{HNO}}_{\text{3}}\right)\text{=}\text{15}\text{.9M}\\ M_1\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{=}\text{18M}\\ M_1\left({\text{CH}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)\text{=}\text{17}\text{.5M}\\ M_1\left({\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right)\text{=}\text{14}\text{.9M}\end{array}$

Calculation:

The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

For $\left(\text{HCl}\right)$ :

By substituting the given values in the formula and the initial volume is,

  $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{\left(\text{3}\text{.0M}\right)\text{×}\left(\text{225mL}\right)}{\left(\text{12}\text{.1M}\right)}\\ \text{=}\text{55}\text{.78mL}\end{array}$

For $\left({\text{HNO}}_{\text{3}}\right)$ :

By substituting the given values in the formula and the initial volume is,

  $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{\left(\text{3}\text{.0M}\right)\text{×}\left(\text{225mL}\right)}{\left(\text{15}\text{.9M}\right)}\\ \text{=}\text{42}\text{.45mL}\end{array}$

For $\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$ :

By substituting the given values in the formula and the initial volume is,

  $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_1=\frac{M_2{V}_2}{M_1}\\ V_1\text{=}\frac{\left(\text{3}\text{.0M}\right)\text{×}\left(\text{225mL}\right)}{\left(\text{18M}\right)}\\ \text{=}\text{37}\text{.5mL}\end{array}$

For $\left({\text{CH}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$ :

By substituting the given values in the formula and the initial volume is,

  $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{\left(\text{3}\text{.0M}\right)\text{×}\left(\text{225mL}\right)}{\left(\text{17}\text{.5M}\right)}\\ \text{=}\text{38}\text{.57mL}\end{array}$

For $\left({\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ :

By substituting the given values in the formula and the initial volume is,

  $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{\left(\text{3}\text{.0M}\right)\text{×}\left(\text{225mL}\right)}{\left(\text{14}\text{.9M}\right)}\\ \text{=}\text{45}\text{.3mL}\end{array}$

Conclusion

The initial molarities for the solutions $\text{HCl,}{\text{HNO}}_{\text{3}}\text{,}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{,}{\text{CH}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{,}\text{and}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ are $\text{55}\text{.78mL,}\text{42}\text{.45mL,}\text{38}\text{.57mL}\text{and}\text{45}\text{.3mL}$ respectively.