Chapter 22, Problem 22.89QP

Interpretation Introduction

To determine: The equilibrium concentrations of cysteine and penicillamine in a given reaction.

Interpretation: The equilibrium concentrations of cysteine and penicillamine in a given reaction are to be calculated.

Concept introduction: The equilibrium concentrations are defined as the concentrations of reactants and products in a reaction at the equilibrium conditions.

Answer to Problem 22.89QP

The equilibrium concentration of cysteine is $\underset{\_}{5.6\times 10^{-4}mM}$ and the equilibrium concentration of penicillamine is $\underset{\_}{4.43\times 10^{-4}mM}$ in a given reaction.

Explanation of Solution

The concentration of cysteine is $\text{1}\text{.00}\text{M}$.

The concentration of ${\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}$ is $\text{1}\text{.00}\text{mM}$.

The value of equilibrium constant $K$ is $0.633$.

The given reaction is,

${\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}\left(aq\right)+\text{cysteine}\left(aq\right)\rightleftharpoons {\text{CH}}_{\text{3}}\text{Hg}{\left(\text{cysteine}\right)}^{+}\left(aq\right)+\text{penicillamine}\left(aq\right)$

The equilibrium constant for the above reaction is,

$K=\frac{\left[{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{cysteine}\right)}^{+}\right]\left[\text{penicillamine}\right]}{\left[{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}\right]\left[\text{cysteine}\right]}$ (1)

The ICE table for the above reaction is below.

$\begin{array}{ccccc}& \left[{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}\right]& \left[\text{cysteine}\right]& \left[{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{cysteine}\right)}^{+}\right]& \left[\text{penicillamine}\right]\\ \text{Initial}& \text{1}\text{.00}\text{mM}& \text{1}\text{.00}\text{M}& 0& 0\\ \text{Change}& -x& -x& +x& +x\\ \text{Equilibrium}& \text{1}\text{.00}\text{mM}-x& \text{1}\text{.00}\text{M}-x& x& x\end{array}$

Substitute the value of equilibrium concentrations from the above table and the value of equilibrium constant in the above expression.

$\begin{array}{c}0.633=\frac{{\left(x\right)}^2}{{\left(\text{1}\text{.00}\text{mM}-x\right)}^2}\\ \sqrt{0.633}=\frac{x}{\text{1}\text{.00}\text{mM}-x}\\ 0.7956\times \left(\text{1}\text{.00}\text{mM}-x\right)=x\\ x=\underset{\_}{4.43\times 10^{-4}mM}\end{array}$

Therefore, the equilibrium concentration of penicillamine is $4.43\times 10^{-4}mM$.

The concentration of cysteine is calculated by the expression,

$\left[\text{cysteine}\right]=0.001\text{mM}-x$

Substitute the value of $x$ in the above expression.

$\begin{array}{c}\left[\text{cysteine}\right]=\text{0}\text{.001}\text{mM}-\text{4}{\text{.43×10}}^{-\text{4}}\text{mM}\\ =\underset{\_}{5.67\times 10^{-4}mM}\end{array}$

Therefore, the equilibrium concentration of cysteine is $\underset{\_}{5.6\times 10^{-4}mM}$.

Conclusion

The equilibrium concentration of cysteine is $\underset{\_}{5.6\times 10^{-4}mM}$ and the equilibrium concentration of penicillamine is $\underset{\_}{4.43\times 10^{-4}mM}$ in a given reaction.