Chapter 12, Problem 12.20P

Interpretation Introduction

Interpretation:

Iterations have to be carried to find $\left[{\text{Pb}}^{2+}\right]$ and ionic strength of ${\text{PbF}}_2$ by approximation.

Concept Introduction:

Expression to compute ionic strength is as follows:

  $\text{μ}=\frac{1}{2}{\displaystyle \sum c_i{z}_i^2}$

Here

$z$ denotes charge.

$c_i$ denotes concentration.

$\text{μ}$ denotes ionic strength.

Explanation of Solution

Equilibrium for ${\text{PbF}}_2$ written as follows:

  ${\text{PbF}}_2\rightleftharpoons {\text{Pb}}^{2+}+2{\text{F}}^{-}$

Assume $\left[{\text{Pb}}^{2+}\right]$ is same as $\left[{\text{F}}^{-}\right]$ equal to $x$.

Corresponding $K_{\text{sp}}$ expression in terms of activity coefficients is as written as follows:

  $K_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{F}}^{-}\right]}^2$        (1)

Substitute $3.6\times {\text{10}}^{-8}$ for $K_{\text{sp}}$ , $x$ for $\left[{\text{Pb}}^{2+}\right]$ and $2x$ for $\left[{\text{F}}^{-}\right]$ in equation (1).

  $\begin{array}{c}3.6\times {\text{10}}^{-8}=4x^3\\ x=2.08\times {\text{10}}^{-3}\text{ M}\end{array}$

Thus when ionic strength is assumed as zero, the value for $\left[{\text{Pb}}^{2+}\right]$ is $2.08\times {\text{10}}^{-3}\text{ M}$.

Formula to compute ionic strength is as follows:

  $\text{μ}=\frac{1}{2}\left(c_1{z}_1^2+c_2{z}_2^2\right)$        (2)

Substitute $+2$ for $z_1$, $2.08\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-1$ for $z_2$, $4.16\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(2.08\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(4.16\times {\text{10}}^{-3}\text{ M}\right){\left(-1\right)}^2\right)\\ =6.24\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.00624 M}\end{array}$

Expression for linear interpolation is given as follows:

  $\frac{\text{unknown γ interval}}{\Delta \text{γ}}=\frac{\text{unknown μ interval}}{\Delta \text{μ}}$        (3)

Substitute $0.665-0.742$ for $\Delta \text{γ}$, $0.665-\text{γ}$ for unknown $\text{γ}$ interval, $0.01-0.005$ for $\Delta \text{μ}$ and $0.01-0.00624$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.665-\text{γ}}{0.665-0.742}=\frac{0.01-0.00624}{0.01-0.005}\\ \text{γ}=0.723\end{array}$

Thus activity coefficient of ${\text{Pb}}^{2+}$  ion is 0.723.

Substitute $0.900-0.926$ for $\Delta \text{γ}$, $0.900-\text{γ}$ for unknown $\text{γ}$ interval, $0.01-0.005$ for $\Delta \text{μ}$ and $0.01-0.00624$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{F}}^{-}$ ion.

  $\begin{array}{c}\frac{0.900-\text{γ}}{0.900-0.926}=\frac{0.01-0.00624}{0.01-0.005}\\ \text{γ}=0.919\end{array}$

Thus activity coefficient of ${\text{F}}^{-}$  ion is 0.919.

Corresponding $K_{\text{sp}}$ expression in terms of activity coefficients is as written as follows:

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

Substitute 0.723 for ${\text{γ}}_{{\text{Pb}}^{2+}}$, 0.919 for ${\text{γ}}_{{\text{F}}^{-}}$,  and $3.6\times {\text{10}}^{-8}$ for $K_{\text{sp}}$ and $x$ for $\left[{\text{Pb}}^{2+}\right]$ and $2x$ for $\left[{\text{F}}^{-}\right]$ in equation (4).

  $\begin{array}{c}3.6\times {\text{10}}^{-8}=4x^3\left(0.723\right){\left(0.919\right)}^2\\ x=2.45\times {\text{10}}^{-3}\text{ M}\end{array}$

Thus, value of $\left[{\text{Pb}}^{2+}\right]$ in this iteration is $2.45\times {\text{10}}^{-3}\text{ M}$ while value of $\left[{\text{F}}^{-}\right]$ is $4.9\times {\text{10}}^{-3}\text{ M}$.

For the next iteration,

Substitute $+2$ for $z_1$, $2.45\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-1$ for $z_2$, $4.9\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(2.45\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(4.9\times {\text{10}}^{-3}\text{ M}\right){\left(-1\right)}^2\right)\\ =7.35\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.00735 M}\end{array}$

For $\text{μ}$ that includes $\text{0}\text{.00735 M}$ interval for $\text{γ}$ is from 0.005 to 0.01.

Substitute $0.665-0.742$ for $\Delta \text{γ}$, $0.665-\text{γ}$ for unknown $\text{γ}$ interval, $0.01-0.005$ for $\Delta \text{μ}$ and $0.01-0.00735$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.665-\text{γ}}{0.665-0.742}=\frac{0.01-0.00735}{0.01-0.005}\\ \text{γ}=0.705\end{array}$

Thus activity coefficient of ${\text{Pb}}^{2+}$  ion is 0.705.

Substitute $0.900-0.926$ for $\Delta \text{γ}$, $0.900-\text{γ}$ for unknown $\text{γ}$ interval, $0.01-0.005$ for $\Delta \text{μ}$ and $0.01-0.00735$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{F}}^{-}$ ion.

  $\begin{array}{c}\frac{0.900-\text{γ}}{0.900-0.926}=\frac{0.01-0.00735}{0.01-0.005}\\ \text{γ}=0.914\end{array}$

Thus activity coefficient of ${\text{F}}^{-}$  ion is 0.914.

Substitute 0.705 for ${\text{γ}}_{{\text{Pb}}^{2+}}$, 0.914 for ${\text{γ}}_{{\text{F}}^{-}}$,  and $3.6\times {\text{10}}^{-8}$ for $K_{\text{sp}}$ and $x$ for $\left[{\text{Pb}}^{2+}\right]$ and $2x$ for $\left[{\text{F}}^{-}\right]$ in equation (4).

  $\begin{array}{c}3.6\times {\text{10}}^{-8}=4x^3\left(0.705\right){\left(0.914\right)}^2\\ x=2.48\times {\text{10}}^{-3}\text{ M}\end{array}$

Thus, value of $\left[{\text{Pb}}^{2+}\right]$ in this iteration is $2.48\times {\text{10}}^{-3}\text{ M}$ while value of $\left[{\text{F}}^{-}\right]$ is $4.96\times {\text{10}}^{-3}\text{ M}$.

For the next iteration,

Substitute $+2$ for $z_1$, $2.48\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-1$ for $z_2$, $4.96\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(2.48\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(4.96\times {\text{10}}^{-3}\text{ M}\right){\left(-1\right)}^2\right)\\ =7.44\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.00744 M}\end{array}$

For $\text{μ}$ that includes $\text{0}\text{.00744 M}$ interval for $\text{γ}$ is from 0.005 to 0.01.

Substitute $0.665-0.742$ for $\Delta \text{γ}$, $0.665-\text{γ}$ for unknown $\text{γ}$ interval, $0.01-0.005$ for $\Delta \text{μ}$ and $0.01-0.00744$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.665-\text{γ}}{0.665-0.742}=\frac{0.01-0.00744}{0.01-0.005}\\ \text{γ}=0.704\end{array}$

Thus activity coefficient of ${\text{Pb}}^{2+}$  ion is 0.704.

Substitute $0.900-0.926$ for $\Delta \text{γ}$, $0.900-\text{γ}$ for unknown $\text{γ}$ interval, $0.01-0.005$ for $\Delta \text{μ}$ and $0.01-0.00744$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{F}}^{-}$ ion.

  $\begin{array}{c}\frac{0.900-\text{γ}}{0.900-0.926}=\frac{0.01-0.00744}{0.01-0.005}\\ \text{γ}=0.913\end{array}$

Thus activity coefficient of ${\text{F}}^{-}$  ion is 0.913.

Substitute 0.704 for ${\text{γ}}_{{\text{Ca}}^{2+}}$, 0.913 for ${\text{γ}}_{{\text{F}}^{-}}$,  and $3.6\times {\text{10}}^{-8}$ for $K_{\text{sp}}$ and $x$ for $\left[{\text{Pb}}^{2+}\right]$ and $2x$ for $\left[{\text{F}}^{-}\right]$ in equation (4).

  $\begin{array}{c}3.6\times {\text{10}}^{-8}=4x^3\left(0.704\right){\left(0.913\right)}^2\\ x=2.48\times {\text{10}}^{-3}\text{ M}\end{array}$

At this stage no more iteration is required as concentration has not changed.