Chapter 16, Problem 62AE

(a)

Interpretation Introduction

Interpretation:

Whether mixture of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ forms precipitate or not has to be calculated.

Concept Introduction:

As two substances get mixed, their molarity changes and volume gets added up. Thus, formula used to calculate new concentration of different substances is as follows:

  $M_{\text{i}}{V}_{\text{i}}=M_{\text{v}}{V}_{\text{v}}$

Here,

$M_{\text{i}}$ is initial molarity

$V_{\text{i}}$ is initial volume

$M_{\text{v}}$ is final molarity

$V_{\text{v}}$ is final volume

Equilibrium quotient is denoted with $\text{Q}$ and is product of concentration of ions raised to proper power of new molecule formed by mixture of two molecules. It explains about two major points that are as follows:

  $\begin{array}{l}\text{Q}>K_{s\text{p}}\text{ precipitation occurs}\\ \text{Q}<K_{s\text{p}}\text{ no precipitation}\end{array}$

Explanation of Solution

The equation of mixture of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows:

  ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{PbSO}}_{\text{4}}+{\text{NaNO}}_{\text{3}}$

The formula used to calculate new molarity of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is as follows:

  $M_1{V}_1=M_2{V}_2$        (1)

Here,

$M_1$ is initial molarity of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

$V_1$ is initial volume of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

$M_2$ is final molarity of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

$V_2$ is final volume of mixture

Rearrange equation (1) for $M_2$.

  $M_2=\frac{M_1{V}_1}{V_2}$        (2)

Substitute $0.010M$ for $M_1$, $\text{100 mL}$ for $V_1$, and $\text{200 mL}$ for $V_2$ in equation (2).

  $\begin{array}{c}{M}_2=\frac{\left(0.010M\right)\left(\text{100 mL}\right)}{\text{200 mL}}\\ =5.0\times {10}^{-3}M\end{array}$

Hence, final molarity of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $5.0\times {10}^{-3}M$, it means the concentration of two ${\text{Na}}^{+}$ is $10.0\times {10}^{-3}M$ and ${\text{SO}}_4^{2-}$ is $5.0\times {10}^{-3}M$.

The formula used to calculate new molarity of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows:

  $M_3{V}_3=M_4{V}_4$        (3)

Here,

$M_3$ is initial molarity of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$V_3$ is initial volume of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$M_4$ is final molarity of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$V_4$ is final volume of mixture

Rearrange equation (1) for $M_4$.

  $M_4=\frac{M_3{V}_3}{V_4}$        (4)

Substitute $0.001M$ for $M_3$, $\text{100 mL}$ for $V_3$, and $\text{200 mL}$ for $V_4$ in equation (4).

  $\begin{array}{c}{M}_4=\frac{\left(0.001M\right)\left(\text{100 mL}\right)}{\text{200 mL}}\\ =5.0\times {10}^{-4}M\end{array}$

Hence, final molarity of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $5.0\times {10}^{-4}M$, it means concentration of ${\text{Pb}}^{+2}$ is and ${\text{NO}}_3^{-}$ is $5.0\times {10}^{-4}M$.

The ionization equation of ${\text{PbSO}}_{\text{4}}$ is as follows:

  ${\text{PbSO}}_{\text{4}}\left(s\right)\rightleftharpoons {\text{Pb}}^{+2}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

The equilibrium quotient for ${\text{PbSO}}_{\text{4}}$ is as follows:

  $\text{Q}=\left[{\text{Pb}}^{+2}\right]\left[{\text{SO}}_4^{2-}\right]$        (5)

Substitute $5.0\times {10}^{-4}M$ for $\left[{\text{Pb}}^{+2}\right]$ and $5.0\times {10}^{-3}M$ for $\left[{\text{SO}}_4^{2-}\right]$ in equation (5).

  $\begin{array}{c}\text{Q}=\left[5.0\times {10}^{-4}\right]\left[5.0\times {10}^{-3}M\right]\\ =25\times {10}^{-7}\end{array}$

For ${\text{PbSO}}_{\text{4}}$ value of $\text{Q}$ is $25\times {10}^{-7}$ and $K_{s\text{p}}$ is $0.13\times {10}^{-7}$. Since $\text{Q}>K_{s\text{p}}$ therefore, precipitation occurs.

(b)

Interpretation Introduction

Interpretation:

Whether mixture of ${\text{AgNO}}_{\text{3}}$ and $\text{NaCl}$ forms precipitate or not has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The equation of mixture of ${\text{AgNO}}_{\text{3}}$ and $\text{NaCl}$ is as follows:

  ${\text{AgNO}}_{\text{3}}+\text{NaCl}\to \text{AgCl}+{\text{NaNO}}_{\text{3}}$

The formula used to calculate new molarity of ${\text{AgNO}}_{\text{3}}$ is as follows:

  $M_1{V}_1=M_2{V}_2$        (1)

Here,

$M_1$ is initial molarity of ${\text{AgNO}}_{\text{3}}$

$V_1$ is initial volume of ${\text{AgNO}}_{\text{3}}$

$M_2$ is final molarity of ${\text{AgNO}}_{\text{3}}$

$V_2$ is final volume of mixture

Rearrange equation (1) for $M_2$.

  $M_2=\frac{M_1{V}_1}{V_2}$        (2)

Substitute $1\times {10}^{-4}M$ for $M_1$, $\text{50 mL}$ for $V_1$, and $\text{150 mL}$ for $V_2$ in equation (2).

  $\begin{array}{c}{M}_2=\frac{\left(1\times {10}^{-4}M\right)\left(\text{50 mL}\right)}{\text{150 mL}}\\ =0.33\times {10}^{-4}M\end{array}$

Hence, final molarity of ${\text{AgNO}}_{\text{3}}$ is $0.33\times {10}^{-4}M$, it means the concentration of ${\text{Ag}}^{+}$ and ${\text{NO}}_3^{-}$ is $0.33\times {10}^{-4}M$.

The formula used to calculate new molarity of $\text{NaCl}$ is as follows:

  $M_3{V}_3=M_4{V}_4$        (3)

Here,

$M_3$ is initial molarity of $\text{NaCl}$

$V_3$ is initial volume of $\text{NaCl}$

$M_4$ is final molarity of $\text{NaCl}$

$V_4$ is final volume of mixture

Rearrange equation (1) for $M_4$.

  $M_4=\frac{M_3{V}_3}{V_4}$        (4)

Substitute $1\times {10}^{-4}M$ for $M_3$, $\text{100 mL}$ for $V_3$, and $\text{150 mL}$ for $V_4$ in equation (4).

  $\begin{array}{c}{M}_4=\frac{\left(1\times {10}^{-4}M\right)\left(\text{100 mL}\right)}{\text{150 mL}}\\ =0.66\times {10}^{-4}M\end{array}$

Hence, final molarity of $\text{NaCl}$ is $0.66\times {10}^{-4}M$, it means the concentration of ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ is $0.66\times {10}^{-4}M$.

The ionization equation of $\text{AgCl}$ is as follows:

  $\text{AgCl}\left(s\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

The equilibrium quotient for $\text{AgCl}$ is as follows:

  $\text{Q}=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]$        (5)

Substitute $0.33\times {10}^{-4}$ for $\left[{\text{Ag}}^{+}\right]$ and $0.66\times {10}^{-4}$ for $\left[{\text{Cl}}^{-}\right]$ in equation (5).

  $\begin{array}{c}\text{Q}=\left[0.33\times {10}^{-4}\right]\left[0.66\times {10}^{-4}\right]\\ =22.2\times {10}^{-10}\end{array}$

For $\text{AgCl}$ value of $\text{Q}$ is $22.2\times {10}^{-10}$ and $K_{s\text{p}}$ is $1.7\times {10}^{-10}$. Since $\text{Q}>K_{s\text{p}}$ therefore, precipitation occurs.

(c)

Interpretation Introduction

Interpretation:

Whether mixture of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{NaOH}$ forms precipitate or not has to be calculated.

Concept Introduction:

Concentration of solution is expressed in terms of molarity. It is the ratio of moles of solute to the volume of solution in litres. The expression used to determine the concentration of solution is as follows:

  $\text{Molarity of solution}=\frac{\text{moles of solute}}{\text{volume of solution}\left(\text{L}\right)}$

Mole is the unit that measures the quantity such as atoms, volume and molecules. It is the ratio of given mass to molecular mass of substance. It is expressed as follows:

  $\text{Number of moles}=\frac{\text{given mass}}{\text{molecular mass}}$

Explanation of Solution

The equation of mixture of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{NaOH}$ is as follows:

  $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}+2\text{NaOH}\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}+2{\text{NaNO}}_{\text{3}}$

Moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\text{Number of moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\text{ mass of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{molecular mass of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$        (6)

Substitute $1\text{ g}$ for mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $164.08\text{ g/mol}$ for molecular mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in equation (6).

  $\begin{array}{c}\text{Number of moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\text{1 g}}{\text{164}\text{.08 g/mol}}\\ =0.0061\text{ mol}\end{array}$

Molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\text{moles of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{volume of solution}\left(\text{L}\right)}$        (7)

Substitute $0.0061\text{ mol}$ for moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{400 mL}$ for volume of solution in equation (7).

$\begin{array}{c}\text{Molarity of Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\left(\frac{\text{0}\text{.0061 mol}}{\text{400 mL}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\\ =2\times {10}^{-8}\text{ mol/L}\end{array}$

The formula used to calculate new molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows:

  $M_1{V}_1=M_2{V}_2$        (1)

Here,

$M_1$ is initial molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$V_1$ is initial volume of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$M_2$ is final molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$V_2$ is final volume of mixture

Rearrange equation (1) for $M_2$.

  $M_2=\frac{M_1{V}_1}{V_2}$        (2)

Substitute $2\times {10}^{-8}M$ for $M_1$, $\text{150 mL}$ for $V_1$, and $\text{400 mL}$ for $V_2$ in equation (2).

  $\begin{array}{c}{M}_2=\frac{\left(2\times {10}^{-8}M\right)\left(\text{150 mL}\right)}{\text{400 mL}}\\ =5.71\times {10}^{-9}M\end{array}$

Hence, final molarity of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $5.71\times {10}^{-9}M$, it means the concentration of ${\text{Ca}}^{+2}$ is $5.71\times {10}^{-9}M$.

The formula used to calculate new molarity of $\text{NaOH}$ is as follows:

  $M_3{V}_3=M_4{V}_4$        (3)

Here,

$M_3$ is initial molarity of $\text{NaOH}$

$V_3$ is initial volume of $\text{NaOH}$

$M_4$ is final molarity of $\text{NaOH}$

$V_4$ is final volume of mixture

Rearrange equation (1) for $M_4$.

  $M_4=\frac{M_3{V}_3}{V_4}$        (4)

Substitute $0.01M$ for $M_3$, $\text{250 mL}$ for $V_3$, and $\text{400 mL}$ for $V_4$ in equation (4).

00

  $\begin{array}{c}{M}_4=\frac{\left(0.01M\right)\left(\text{250 mL}\right)}{\text{400 mL}}\\ =6.25\times {10}^{-3}M\end{array}$

Hence, final molarity of $\text{NaOH}$ is $6.25\times {10}^{-3}M$, it means the concentration of ${\text{Na}}^{+}$ and ${\text{OH}}^{-}$ is $6.25\times {10}^{-3}M$.

The ionization equation of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

  $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftharpoons {\text{Ca}}^{+2}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

The equilibrium quotient for $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

  $\text{Q}=\left[{\text{Ca}}^{+2}\right]{\left[{\text{OH}}^{-}\right]}^2$        (5)

Substitute $5.71\times {10}^{-9}$ for $\left[{\text{Ca}}^{+2}\right]$ and $12.5\times {10}^{-3}$ for $\left[{\text{OH}}^{-}\right]$ in equation (5).

  $\begin{array}{c}\text{Q}=\left[5.71\times {10}^{-9}\right]{\left[12.5\times {10}^{-3}\right]}^2\\ =8.92\times {10}^{-13}\end{array}$

For $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ value of $\text{Q}$ is $8.92\times {10}^{-13}$ and $K_{s\text{p}}$ is $1.3\times {10}^{-6}$. Since $\text{Q}<K_{s\text{p}}$ therefore,  no precipitation occurs.