Chapter 12, Problem 12.21P

(a)

Interpretation Introduction

Interpretation:

Iterations have to be carried to find values of $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ until concentration does not changes.

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 ${\text{CaSO}}_4$ written as follows:

  ${\text{CaSO}}_4\rightleftharpoons {\text{Ca}}^{2+}+{\text{SO}}_4^{2-}$

Assume $\left[{\text{Ca}}^{2+}\right]$ is same as $\left[{\text{SO}}_4^{2-}\right]$ equal to $x$.

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

  $K_{\text{sp}}=\left[{\text{Ca}}^{2+}\right]\left[{\text{SO}}_4^{2-}\right]$        (1)

Substitute $2.4\times {\text{10}}^{-5}$ for $K_{\text{sp}}$ and $x$ for both $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ in equation (1).

  $\begin{array}{c}2.4\times {\text{10}}^{-5}=x^2\\ x=4.8989\times {\text{10}}^{-3}\text{ M}\end{array}$

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 $z_1$, $4.8989\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-2$ for $z_2$, $4.8989\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(4.8989\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(4.8989\times {\text{10}}^{-3}\text{ M}\right){\left(-2\right)}^2\right)\\ =19.6\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.0196 M}\end{array}$

For ${\text{Ca}}^{2+}$ ion, interval for $\text{μ}$ that includes $\text{0}\text{.0196 M}$ is 0.485 to 0.675 and corresponding interval for $\text{γ}$ is from 0.05 to 0.01.

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 $0.485-0.675$ for $\Delta \text{γ}$, $0.485-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.0196$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.485-\text{γ}}{0.485-0.675}=\frac{0.05-0.0196}{0.05-0.01}\\ \text{γ}=0.629\end{array}$

Thus activity coefficient of ${\text{Ca}}^{2+}$  ion is 0.629.

Substitute $0.445-0.660$ for $\Delta \text{γ}$, $0.445-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.0196$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{SO}}_4^{2-}$ ion.

  $\begin{array}{c}\frac{0.445-\text{γ}}{0.445-0.660}=\frac{0.05-0.0196}{0.05-0.01}\\ \text{γ}=0.6084\end{array}$

Thus activity coefficient of ${\text{SO}}_4^{2-}$  ion is 0.6084.

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

  $K_{\text{sp}}=\left[{\text{Ca}}^{2+}\right]\left({\text{γ}}_{{\text{Ca}}^{2+}}\right)\left[{\text{SO}}_4^{2-}\right]\left({\text{γ}}_{{\text{SO}}_4^{2-}}\right)$        (4)

Substitute 0.629 for ${\text{γ}}_{{\text{Ca}}^{2+}}$, 0.608 for ${\text{γ}}_{{\text{SO}}_4^{2-}}$,  and $2.4\times {\text{10}}^{-5}$ for $K_{\text{sp}}$ and $x$ for both $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ in equation (4).

  $\begin{array}{c}2.4\times {\text{10}}^{-5}=x^2\left(0.629\right)\left(0.608\right)\\ x=7.92\times {\text{10}}^{-3}\text{ M}\end{array}$

For the next iteration,

Substitute $+2$ for $z_1$, $7.92\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-2$ for $z_2$, $7.92\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(7.92\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(7.92\times {\text{10}}^{-3}\text{ M}\right){\left(-2\right)}^2\right)\\ =31.7\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.0317 M}\end{array}$

For ${\text{Ca}}^{2+}$ ion, interval for $\text{μ}$ that includes $\text{0}\text{.0317 M}$ is 0.485 to 0.675 and corresponding interval for $\text{γ}$ is from 0.05 to 0.01.

Substitute $0.485-0.675$ for $\Delta \text{γ}$, $0.485-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.0317$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.485-\text{γ}}{0.485-0.675}=\frac{0.05-0.0317}{0.05-0.01}\\ \text{γ}=0.572\end{array}$

Thus activity coefficient of ${\text{Ca}}^{2+}$  ion is 0.572.

Substitute $0.445-0.660$ for $\Delta \text{γ}$, $0.445-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.0317$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{SO}}_4^{2-}$ ion.

  $\begin{array}{c}\frac{0.445-\text{γ}}{0.445-0.660}=\frac{0.05-0.0317}{0.05-0.01}\\ \text{γ}=0.543\end{array}$

Thus activity coefficient of ${\text{SO}}_4^{2-}$  ion is 0.543.

Substitute 0.572 for ${\text{γ}}_{{\text{Ca}}^{2+}}$, 0.543 for ${\text{γ}}_{{\text{SO}}_4^{2-}}$,  and $2.4\times {\text{10}}^{-5}$ for $K_{\text{sp}}$ and $x$ for both $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ in equation (4).

  $\begin{array}{c}2.4\times {\text{10}}^{-5}=x^2\left(0.572\right)\left(0.573\right)\\ x=8.79\times {\text{10}}^{-3}\text{ M}\end{array}$

For the next iteration,

Substitute $+2$ for $z_1$, $8.79\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-2$ for $z_2$, $8.79\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(8.79\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(8.79\times {\text{10}}^{-3}\text{ M}\right){\left(-2\right)}^2\right)\\ =35.16\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.03516 M}\end{array}$

For ${\text{Ca}}^{2+}$ ion, interval for $\text{μ}$ that includes $\text{0}\text{.03516 M}$ is 0.485 to 0.675 and corresponding interval for $\text{γ}$ is from 0.05 to 0.01.

Substitute $0.485-0.675$ for $\Delta \text{γ}$, $0.485-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.03516$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.485-\text{γ}}{0.485-0.675}=\frac{0.05-0.0317}{0.05-0.01}\\ \text{γ}=0.555\end{array}$

Thus activity coefficient of ${\text{Ca}}^{2+}$  ion is 0.572.

Substitute $0.445-0.660$ for $\Delta \text{γ}$, $0.445-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.03516$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{SO}}_4^{2-}$ ion.

  $\begin{array}{c}\frac{0.445-\text{γ}}{0.445-0.660}=\frac{0.05-0.0317}{0.05-0.01}\\ \text{γ}=0.525\end{array}$

Thus activity coefficient of ${\text{SO}}_4^{2-}$  ion is 0.525.

Substitute 0.555 for ${\text{γ}}_{{\text{Ca}}^{2+}}$, 0.525 for ${\text{γ}}_{{\text{SO}}_4^{2-}}$,  and $2.4\times {\text{10}}^{-5}$ for $K_{\text{sp}}$ and $x$ for both $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ in equation (4).

  $\begin{array}{c}2.4\times {\text{10}}^{-5}=x^2\left(0.555\right)\left(0.525\right)\\ x=9.08\times {\text{10}}^{-3}\text{ M}\end{array}$

For the next iteration,

Substitute $+2$ for $z_1$, $9.08\times {\text{10}}^{-3}\text{ M}$ for $c_1$, $-2$ for $z_2$, $9.08\times {\text{10}}^{-3}\text{ M}$ for $c_2$ in equation (2).

  $\begin{array}{c}\text{μ}=\frac{1}{2}\left(\left(9.08\times {\text{10}}^{-3}\text{ M}\right){\left(+2\right)}^2+\left(9.08\times {\text{10}}^{-3}\text{ M}\right){\left(-2\right)}^2\right)\\ =36.32\times {10}^{-3}\text{ M}\\ \approx \text{0}\text{.0363 M}\end{array}$

For ${\text{Ca}}^{2+}$ ion, interval for $\text{μ}$ that includes $\text{0}\text{.0363 M}$ is 0.485 to 0.675 and corresponding interval for $\text{γ}$ is from 0.05 to 0.01.

Substitute $0.485-0.675$ for $\Delta \text{γ}$, $0.485-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.0363$ for known $\text{μ}$ interval equation (3).

  $\begin{array}{c}\frac{0.485-\text{γ}}{0.485-0.675}=\frac{0.05-0.0363}{0.05-0.01}\\ \text{γ}=0.550\end{array}$

Thus activity coefficient of ${\text{Ca}}^{2+}$  ion is 0.550.

Substitute $0.445-0.660$ for $\Delta \text{γ}$, $0.445-\text{γ}$ for unknown $\text{γ}$ interval, $0.05-0.01$ for $\Delta \text{μ}$ and $0.05-0.0363$ for known $\text{μ}$ interval equation (3) to calculate $\text{γ}$ for ${\text{SO}}_4^{2-}$ ion.

  $\begin{array}{c}\frac{0.445-\text{γ}}{0.445-0.660}=\frac{0.05-0.0363}{0.05-0.01}\\ \text{γ}=0.519\end{array}$

Thus activity coefficient of ${\text{SO}}_4^{2-}$  ion is 0.519.

Substitute 0.550 for ${\text{γ}}_{{\text{Ca}}^{2+}}$, 0.519 for ${\text{γ}}_{{\text{SO}}_4^{2-}}$,  and $2.4\times {\text{10}}^{-5}$ for $K_{\text{sp}}$ and $x$ for both $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ in equation (4).

  $\begin{array}{c}2.4\times {\text{10}}^{-5}=x^2\left(0.550\right)\left(0.519\right)\\ x=9.16\times {\text{10}}^{-3}\text{ M}\\ \approx \text{9}\text{.2 mM}\end{array}$

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

(b)

Interpretation Introduction

Interpretation:

Iterations have to be carried to find values of $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ until concentration does not changes.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Equilibrium for ${\text{CaSO}}_4$ is written as follows:

  ${\text{CaSO}}_4\rightleftharpoons {\text{Ca}}^{2+}+{\text{SO}}_4^{2-}$

Concentration of $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ are $\text{9}\text{.2 mM}$ and rest of the millimoles concentration can be attributed to formation of ion pair ${\text{CaSO}}_4\left(aq\right)$ as ${\text{CaSO}}_4$ is only sparingly soluble salt.