Chapter 15, Problem 33A

Interpretation Introduction

Interpretation: The volume of the solution which required to prepare a 3.0M solution, is to be calculated.

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 33A

The required volume of the solution is $\text{6}\text{.21 mL}$ .

Explanation of Solution

Given information:

The initial volume of the solution is, $V_1=1\text{25 mL}$

The initial molarity and final molarity of the solution is $M_1=0.150M$ and $M_2=3.0\text{2M}$ respectively.

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,

  $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_2=\frac{M_1{V}_1}{M_2}\\ =\frac{\text{0}\text{.150M×125mL}}{\text{3}\text{.02M}}\\ \text{=}\text{6}\text{.21mL}\end{array}$

So measure exactly $\text{6}\text{.21mL}$ of the $3.0\text{2M}$ sodium hydroxide and make to exactly $\text{125mL}$ .

But it is very complicated to make $\text{125mL}$ solution because 12mL volumetric flask is not available there.

Therefore, student can add $\text{12}\text{.50mL}$ of $\text{3}\text{.02M NaOH}$ to a $\text{250mL}$ volumetric flask , and then added the water to the mark.

Conclusion

So measure exactly $6.2\text{1mL}$ of the $\text{3}\text{.02M}$ sodium hydroxide and make to exactly $\text{125mL}$ .