Chapter 4, Problem 4.91QE

(a)

Interpretation Introduction

Interpretation:

The volume of ${\text{HNO}}_{\text{3}}$ has to be calculated.

Concept Introduction:

Titration: It is a quantitative analysis method used to express the concentration of acid or base present in the solution.

For an acid-base titration ${\left(Eq\right)}_{acid}={\left(Eq\right)}_{base}$

That is,

  $N_{acid}\times V_{acid}=N_{base}\times V_{base}$

  Here, V is volume and N is normality.

Normality: It generally expresses the concentration of acid or base as equivalents of acid or base present in one liter of the solution.

Molarity: The concentration for solutions is expressed in terms of molarity as follows,

  $\text{M = }\frac{\text{n}}{V}$

  Here, n is number of solute moles and V is volume of solution in liters.

The normality for any acid or base equals the molarity times of the number of ${\text{H}}^{\text{+}}\text{and}{\text{OH}}^{\text{-}}$ correspondingly.

Equivalent of Acid: Generally, 1 equivalent of ion is the number of ions which has charge of one mole.

Answer to Problem 4.91QE

The volume of ${\text{HNO}}_{\text{3}}$ is $15\text{ mL}$.

Explanation of Solution

Given:

  $\begin{array}{l}{\text{M}}_{{\text{HNO}}_{\text{3}}}\text{=}0.22\text{3M}\\ {\text{M}}_{\text{Ba}{\left(OH\right)}_2}\text{=}\text{0}\text{.033M}\\ {\text{V}}_{\text{Ba}{\left(OH\right)}_2}\text{= 50}\text{.00mL}\\ {\text{V}}_{{\text{HNO}}_{\text{3}}}\text{=}\text{?}\end{array}$

First, the concentration of ${\text{OH}}^{\text{-}}$ ions present in $Ba{\left(OH\right)}_2$ solution is as follows,

    $\begin{array}{c}\text{Molarity = }\frac{\text{Solute moles}}{\text{Volume of solution in L}}\\ \\ 0.033\text{M}=\frac{\text{Solute moles}}{\text{50}\times {\text{10}}^{\text{-3}}\text{ L}}\\ \\ \text{Solute moles}=0.033\text{M}\times \text{50}\times {\text{10}}^{\text{-3}}\text{L }=1.65\times {10}^{-3}\text{ moles}\\ \\ {\text{OH}}^{-}=1.65\times {10}^{-3}\text{ moles}\end{array}$

The equation for given reaction is $2HN{O}_3+Ba{\left(OH\right)}_2\to Ba{\left(N{O}_3\right)}_2+2H_{2}O$ shows that two moles of acid requires one mole of base hence they are in $\text{2:1}$ ratio.

The mole of ${\text{H}}^{\text{+}}$ required is equal to the two times of the ${\text{OH}}^{\text{-}}$ ions value that is $3.3\times {10}^{-3}$. Therefore the volume of acid is as follows,

    $\begin{array}{c}\text{Molarity = }\frac{\text{Solute moles}}{\text{Volume of solution in L}}\\ \\ \text{0}\text{.223M}=\frac{\text{3}\text{.3}\times {\text{10}}^{\text{-3}}}{\text{volume}}\\ \\ {\text{volume of HNO}}_3=0.0147\text{ L}\\ \\ \text{= 15mL}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The volume of ${\text{AgNO}}_{\text{3}}$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 4.91QE

The volume of ${\text{AgNO}}_{\text{3}}$ is $43\text{ mL}$.

Explanation of Solution

Given:

  $\begin{array}{l}{\text{M}}_{{\text{AgNO}}_{\text{3}}}\text{=}1.13\text{M}\\ {\text{M}}_{C{l}^{-}}\text{=}\text{2}\text{.43M}\\ {\text{V}}_{C{l}^{-}}\text{= 10}\text{.00mL}\\ {\text{V}}_{{\text{AgNO}}_{\text{3}}}\text{=}\text{?}\end{array}$

First, the concentration of ${\text{Cl}}^{\text{-}}$ ions present is calculated as follows,

    $\begin{array}{c}\text{Molarity = }\frac{\text{Solute moles}}{\text{Volume of solution in L}}\\ \\ 2.43\text{M}=\frac{{\text{Cl}}^{\text{-}}\text{ moles}}{\text{10}\times {\text{10}}^{\text{-3}}\text{ L}}\\ \\ {\text{Cl}}^{\text{-}}\text{ moles}=2.43\text{M}\times 1\text{0}\times {\text{10}}^{\text{-3}}\text{L }=0.0243\text{ moles}\end{array}$

The equation for given reaction is $AgN{O}_3+2C{l}^{-}\to AgC{l}_2+N{O}_3^{-}$ shows that they are in $\text{1:2}$ ratio.

The mole of ${\text{Ag}}^{\text{+}}$ required is equal to the two times the ${\text{Cl}}^{\text{-}}$ ions value that is $0.0486$. Therefore the volume is as follows,

    $\begin{array}{c}\text{Molarity = }\frac{\text{Solute moles}}{\text{Volume of solution in L}}\\ \\ \text{1}\text{.13M}=\frac{0.0486\text{ moles}}{\text{volume}}\\ \\ {\text{volume of AgNO}}_3=0.043\text{ L}\\ \\ \text{= 43 mL}\end{array}$