Chapter 15, Problem 45A

(a)

Interpretation Introduction

Interpretation:

The normality of the given solution of 0.134 M NaOH is to be calculated.

Concept Introduction:

The normality of the solution is given as,

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in litre}}$

Answer to Problem 45A

  $\text{0}\text{.134N}\text{NaOH}$

Explanation of Solution

Given information:

The given solution is, $\text{0}\text{.134 M}\text{NaOH}$

Calculation:

The normality is calculated as follows:

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{volume of}\text{of}\text{solution}\text{in liter}}$

Also, molarity is related to number of moles of solute and volume of solution in liter as follows:

  $\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in liter}}$

Thus,

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Number of moles of solute}}\times \text{molarity}$

Putting the values,

  $\text{0}\text{.134M}\text{NaOH×}\frac{\text{1}\text{equiv}\text{NaOH}}{\text{1}\text{mol}\text{NaOH}}\text{=0}\text{.134N}\text{NaOH}$

Thus, the normality is $\text{0}\text{.134N}\text{NaOH}$ .

(b)

Interpretation Introduction

Interpretation:

The normality of the given solution of $\text{0}\text{.00521M}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is to be calculated.

Concept Introduction:

The normality of the solution is given as,

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in litre}}$

Answer to Problem 45A

  $3.37\times {10}^{-3}\text{N}$

Explanation of Solution

Given information:

The given solution is, $\text{0}\text{.00521M}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$

Calculation:

The normality is calculated as follows:

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{volume of}\text{of}\text{solution}\text{in liter}}$

Also, molarity is related to number of moles of solute and volume of solution in liter as follows:

  $\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in liter}}$

Thus,

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Number of moles of solute}}\times \text{molarity}$

Putting the values,

  $\text{0}\text{.00521M}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{×}\frac{\text{2}\text{equiv}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{1}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}\text{=0}\text{.0104N}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$

Thus, the normality is $\text{0}\text{.0104N}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$

  $\text{0}\text{.0104N}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ .

(c)

Interpretation Introduction

Interpretation:

The normality of the given solution of 4.42 M H₃PO₄ is to be calculated.

Concept Introduction:

The normality of the solution is given as,

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in litre}}$

Answer to Problem 45A

  $\text{13}\text{.3N}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Explanation of Solution

Given information:

The given solution is, $\text{4}\text{.42M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Calculation:

The normality is calculated as follows:

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{volume of}\text{of}\text{solution}\text{in liter}}$

Also, molarity is related to number of moles of solute and volume of solution in liter as follows:

  $\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in liter}}$

Thus,

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Number of moles of solute}}\times \text{molarity}$

Putting the values,

  $\text{4}\text{.42M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{×}\frac{\text{3}\text{equiv}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=13}\text{.3N}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Thus, the normality is $\text{13}{\text{.3 N H}}_{\text{3}}{\text{PO}}_{\text{4}}$ .