Chapter 22, Problem 22.83QP

Interpretation Introduction

Interpretation: The equilibrium concentrations of cysteine and penicillamine for the given data are to be calculated.

Concept introduction: The equilibrium constant $\left(K_{\text{c}}\right)$ is calculated by the formula,

$K_{\text{c}}=\frac{{\left[\text{Product}\right]}^y}{{\left[\text{Reactant}\right]}^x}$

To determine: The equilibrium concentrations of cysteine and penicillamine for the given data.

Answer to Problem 22.83QP

Solution

The equilibrium concentrations of cysteine and penicillamine for the given data are $\underset{\_}{0.9995M}$ and $\underset{\_}{0.0005M}$.

Explanation of Solution

Explanation

Given

The 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 initial concentration of cysteine solution is $1.00\text{M}$.

The initial concentration of ${\text{CH}}_{\text{3}}\text{Hg}{\left(\text{glutathione}\right)}^{+}$ solution is $1.00\text{mM}$.

The equilibrium constant $\left(K_{\text{c}}\right)$ of given reaction is $0.633$.

Change the initial concentration of ${\text{CH}}_{\text{3}}\text{Hg}{\left(\text{glutathione}\right)}^{+}$ in $\text{M}$, use conversion factor.

$\begin{array}{c}1\text{mM}={10}^{-3}\text{M}\\ 1.00\text{mM}=1\times {10}^{-3}\text{M}\\ ={10}^{-3}\text{M}\end{array}$

The equilibrium constant $\left(K_{\text{c}}\right)$ for the given reaction is calculated by the formula,

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

Where,

• $\left[{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{cysteine}\right)}^{+}\left(aq\right)\right]$ is the equilibrium concentration of product ${\text{CH}}_{\text{3}}\text{Hg}{\left(\text{cysteine}\right)}^{+}\left(aq\right)$.
• $\left[\text{penicillamine}\left(aq\right)\right]$ is the equilibrium concentration of product $\text{penicillamine}\left(aq\right)$.
• $\left[{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}\left(aq\right)\right]$ is the equilibrium concentration of reactant ${\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}\left(aq\right)$.
• $\left[\text{cysteine}\left(aq\right)\right]$ is the equilibrium concentration of reactant $\text{cysteine}\left(aq\right)$.

The change in concentration that occurs during the reaction is considered to be $x$.

The table that shows the initial and the equilibrium concentrations of the reactants and products is stated as follows.

$\begin{array}{c}\\ \text{Initial}\\ \text{Equilibrium}\end{array}\begin{array}{ccccccc}{\text{CH}}_{\text{3}}\text{Hg}{\left(\text{penicillamine}\right)}^{+}\left(aq\right)& +& \text{cysteine}\left(aq\right)& \rightleftharpoons & {\text{CH}}_{\text{3}}\text{Hg}{\left(\text{cysteine}\right)}^{+}\left(aq\right)& +& \text{penicillamine}\left(aq\right)\\ {10}^{-3}\text{M}& & 1.00\text{M}& & 0& & 0\\ \left({10}^{-3}-x\right)\text{M}& & \left(1.00-x\right)\text{M}& & x& & x\end{array}$

Substitute the values of equilibrium concentration of reactants, product and equilibrium constant in equation (1).

$\begin{array}{c}0.633=\frac{{\left(x\right)}^2}{\left({10}^{-3}-x\right)\left(1-x\right)}\\ 0.633=\frac{{\left(x\right)}^2}{\left({10}^{-3}-{10}^{-3}x-x+x^2\right)}\\ 0.633=\frac{{\left(x\right)}^2}{\left({10}^{-3}-1.001x+x^2\right)}\\ 0.000633+0.633x+0.633x^2=x^2\end{array}$

Simplify the above equation.

$\begin{array}{c}0.000633+0.633x+0.633x^2=x^2\\ 0.367x^2-0.633x-0.000633=0\end{array}$ (2)

A quadratic equation is solved by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.
• $b$ is the coefficient of $x$.
• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{c}x=\frac{-\left(-0.633\right)\pm \sqrt{{\left(-0.633\right)}^2-4\times 0.367\times 0.000633}}{2\times 0.367}\\ =\frac{0.633\pm 0.632}{0.734}\end{array}$

Simplify, for positive value.

$\begin{array}{c}x=\frac{0.633+0.632}{0.734}\\ =1.723\text{M}\end{array}$ (3)

Simplify, for negative value.

$\begin{array}{c}x=\frac{0.633-0.632}{0.734}\\ =0.0005\text{M}\end{array}$ (4)

The value of $x$ in equation (3) is not possible as the equilibrium concentration of cysteine is $\left(1.00-x\right)\text{M}$.

Thus the equilibrium concentrations penicillamine is $0.0005\text{M}$.

The equilibrium concentration of cysteine from above table is expressed as,

$\text{cysteine}\left(aq\right)=\left(1.00-x\right)\text{M}$ (5)

Substitute the value of $x$ in equation (5).

$\begin{array}{c}\text{cysteine}\left(aq\right)=\left(1.00-0.0005\right)\text{M}\\ =\text{0}\text{.9995}\text{M}\end{array}$

Thus, the equilibrium concentrations of cysteine and penicillamine for the given data are $\underset{\_}{0.9995M}$ and $\underset{\_}{0.0005M}$.

Conclusion

The equilibrium concentrations of cysteine and penicillamine for the given data are $\underset{\_}{0.9995M}$ and $\underset{\_}{0.0005M}$.