Chapter 12, Problem 12.34P

Interpretation Introduction

Interpretation:

Moles of $\text{PbO}$ that will dissolve in $1\text{ L}$ solution with fixed $\text{pH}$ of 10.5 has to be calculated.

Concept Introduction:

Systematic treatment of any chemical equilibrium involves steps indicated as follows:

• Pertinent reactions possible for the system are written.
• Charge balance and mas balance equation are written.
• Equilibrium constant corresponding to equation written are listed.
• Equations are solved to find unknown from available equations.

Explanation of Solution

Formula to calculate $\left[{\text{H}}^{+}\right]$ is given as follows:

  $\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$        (1)

Substitute 10.50 for $\text{pH}$ in equation (1).

  $\left[{\text{H}}^{+}\right]={10}^{-10.50}$

In aqueous solution self-ionization of water occurs as follows:

  ${\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$

Corresponding ionic product expression is written as follows:

  $K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (2)

Rearrange equation (2) to obtain $\left[{\text{OH}}^{-}\right]$.

  $\left[{\text{OH}}^{-}\right]=\frac{K_{\text{w}}}{\left[{\text{H}}^{+}\right]}$        (3)

Equilibrium of $\text{PbO}$ occurs as follows:

  $\text{PbO}\left(s\right)\rightleftharpoons {\text{Pb}}^{2+}+2{\text{OH}}^{-}$

Corresponding solubility product expression is written as follows:

  $K_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$        (4)

Substitute equation (3) in equation (4).

  $K_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]{\left(\frac{K_{\text{w}}}{\left[{\text{H}}^{+}\right]}\right)}^2$        (5)

Substitute ${10}^{-14}$ for $K_{\text{w}}$ , ${10}^{-10.50}$ for $\left[{\text{H}}^{+}\right]$ and $5\times {10}^{-16}$ for $K_{\text{sp}}$ in equation (5) and rearrange.

  $\begin{array}{c}\left[{\text{Pb}}^{2+}\right]=5\times {10}^{-16}{\left(\frac{{10}^{-10.50}}{{10}^{-14}}\right)}^2\\ =5\times {10}^{-9}\end{array}$

Hence, $5\times {10}^{-9}\text{ mol}$ of $\text{PbO}$ can dissolve in $1\text{ L}$ of solution.

Equilibrium reaction for ${\text{PbOH}}^{+}$ occurs as follows:

  ${\text{Pb}}^{2+}+{\text{H}}_2\text{O}\rightleftharpoons {\text{PbOH}}^{+}+{\text{H}}^{+}$

Dissociation constant expression is written as follows:

  $K_{\text{b}}=\frac{\left[\text{HCN}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CN}}^{-}\right]}$        (6)

Substitute $2.5\times {10}^{-8}$ for $K_{\text{a}}$ , ${10}^{-10.50}$ for $\left[{\text{H}}^{+}\right]$ and $5\times {10}^{-9}$ for $\left[{\text{Pb}}^{2+}\right]$ in equation (6) and rearrange.

  $\begin{array}{c}\left[{\text{PbOH}}^{+}\right]=\left(2.5\times {10}^{-8}\right)\left(\frac{5\times {10}^{-9}}{{10}^{-10.50}}\right)\\ =3.95\times {10}^{-6}\\ \approx 4.0\times {10}^{-6}\end{array}$

Thus, $4.0\times {10}^{-6}\text{ mol}$ of ${\text{PbOH}}^{+}$ can dissolve in $1\text{ L}$ of solution.

If activities are included formula to calculate $\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}$ is as follows:

  $\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}=\frac{K_{\text{w}}}{{10}^{-\text{pH}}}$        (7)

Substitute ${10}^{-14}$ for $K_{\text{w}}$ and 10.50 for $\text{pH}$  in equation (7).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}=\frac{{10}^{-14}}{{10}^{-10.50}}\\ ={10}^{-3.50}\end{array}$

Ionic product expression with activities included is written as follows:

  $K_{\text{w}}=\left[{\text{H}}^{+}\right]\left({\text{γ}}_{{\text{H}}^{+}}\right)\left[{\text{OH}}^{-}\right]\left({\text{γ}}_{{\text{OH}}^{-}}\right)$        (2)

Thus, formula to compute $\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}$ is as follows:

  $\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}=\frac{K_{\text{w}}}{{10}^{-\text{pH}}}$        (7)

Substitute ${10}^{-14}$ for $K_{\text{w}}$ and 9 for $\text{pH}$  in equation (7).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}=\frac{{10}^{-14}}{{10}^{-9}}\\ ={10}^{-5}\end{array}$

Dissociation constant expression is written as follows:

  $K_{\text{b}}=\frac{\left[\text{HCN}\right]\left({\text{γ}}_{\text{HCN}}\right)\left[{\text{OH}}^{-}\right]\left({\text{γ}}_{{\text{OH}}^{-}}\right)}{\left[{\text{CN}}^{-}\right]\left({\text{γ}}_{{\text{CN}}^{-}}\right)}$        (6)

Substitute $1.6\times {10}^{-5}$ for $K_{\text{b}}$ and 0.755 for ${\text{γ}}_{{\text{CN}}^{-}}$, 1 for  ${\text{γ}}_{\text{HCN}}$, ${10}^{-5}$ for $\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}$  in equation (6) and rearrange.

  $\begin{array}{c}\left[\text{HCN}\right]=\frac{\left(1.6\times {10}^{-5}\right)\left(0.755\right)\left[{\text{CN}}^{-}\right]}{1.0\times {10}^{-5}}\\ =1.208\left[{\text{CN}}^{-}\right]\end{array}$

Substitute $1.208\left[{\text{CN}}^{-}\right]$ for $\left[\text{HCN}\right]$ in equation ().

  $\begin{array}{l}\left[{\text{Ag}}^{2+}\right]=\left[{\text{CN}}^{-}\right]+1.208\left[{\text{CN}}^{-}\right]\\ \left[{\text{CN}}^{-}\right]=\frac{\left[{\text{Ag}}^{2+}\right]}{2.208}\end{array}$

Substitute  $\left[{\text{Ag}}^{2+}\right]\text{/}20208$ in equation ().

Solubility product expression with activities included is written as follows:

  $K_{\text{sp}}=\left[{\text{Ag}}^{2+}\right]\left({\text{γ}}_{{\text{Ag}}^{2+}}\right)\left[{\text{CN}}^{-}\right]\left({\text{γ}}_{{\text{CN}}^{-}}\right)$        (8)

Substitute 0.455 for ${\text{γ}}_{{\text{Pb}}^{2+}}$, ${10}^{-3.50}$ for $\left[{\text{OH}}^{-}\right]{\text{γ}}_{{\text{OH}}^{-}}$ and $5\times {10}^{-16}$ for $K_{\text{sp}}$ in equation (5).

  $\begin{array}{c}\left[{\text{Pb}}^{2+}\right]=\frac{5\times {10}^{-16}}{\left(0.455\right){\left({10}^{-3.50}\right)}^2}\\ =1.098\times {10}^{-8}\\ \approx 1.1\times {10}^{-8}\text{ M}\end{array}$

Thus, $1.1\times {10}^{-8}\text{ mol}$ of ${\text{Pb}}^{2+}$ can dissolve in $1\text{ L}$ of solution.

Charge balance equation that shows net postive charge equal to net negative charge is formulated as follows:

  $\left[{\text{H}}^{+}\right]\text{+2}\left[{\text{Mg}}^{2+}\right]=\left[{\text{OH}}^{-}\right]$        (3)

Mass balance equation is formulated as follows:

  $\left[{\text{Ag}}^{2+}\right]=\left[{\text{CN}}^{-}\right]+\left[\text{HCN}\right]$        (4)

Substitute equation (4) and equation (2) in equation (3).

  $\left[{\text{H}}^{+}\right]\text{+2}\left(4.0\times {10}^{-8}\text{ M}\right)=\frac{K_{\text{w}}}{\left[{\text{H}}^{+}\right]}$        (5)

Substitute ${10}^{-14}$ for $K_{\text{w}}$ in equation (5) and rearrange.

  ${\left[{\text{H}}^{+}\right]}^2+\left(8\times {10}^{-8}\right)\left[{\text{H}}^{+}\right]-{10}^{-14}=0$

Simplify to obtain value of $\left[{\text{H}}^{+}\right]$ as $6.8\times {10}^{-8}$.

Substitute $6.8\times {10}^{-8}$ for $\left[{\text{H}}^{+}\right]$ in equation (2) to obtain value of $\left[{\text{OH}}^{-}\right]$.

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{6.8\times {10}^{-8}}\\ =1.4\times {10}^{-7}\text{ M}\end{array}$

Hence, values of $\left[{\text{Mg}}^{2+}\right]$, $\left[{\text{OH}}^{-}\right]$ and $\left[{\text{H}}^{+}\right]$ are $4.0\times {10}^{-8}\text{ M}$, $1.4\times {10}^{-7}\text{ M}$ and $6.8\times {10}^{-8}\text{ M}$ respectively.