Chapter 23, Problem 23.84QA

Interpretation Introduction

To find:

The equilibrium concentration of cysteine and glutathione

Answer to Problem 23.84QA

Solution:

[Cysteine] = $3.7 \times {10}^{-4}M$

[Glutathione] = $8.3 \times {10}^{-4}M$

Explanation of Solution

1) Concept:

The equilibrium reaction of ligand exchange of $C{H}_{3}H{g}^{+}$ is

$C{H}_{3}H{g\left(glutathione\right)}^{+} \left(aq\right)+cysteine \left(aq\right) \rightleftharpoons C{H}_{3}H{g\left(cysteine\right)}^{+} \left(aq\right)+glutathione \left(aq\right)$

The initial concentration of cysteine and $C{H}_{3}H{g\left(glutathione\right)}^{+}$ has been provided; therefore, we can calculate the equilibrium concentrations of the reactants and products using RICE table.

2) Formula:

$equilibrium constant K_{ }= \frac{\left[C{H}_{3}H{g\left(cysteine\right)}^{+}\right]\left[glutathione\right]}{\left[C{H}_{3}H{g\left(glutathione\right)}^{+}\right]\left[cysteine\right]}$

3) Given:

i) Initial concentration of cysteine $=1.20 mM=0.0012 M$

ii) Initial concentration of $C{H}_{3}H{g\left(glutathione\right)}^{+}=1.20 mM= 0.0012 M$

iii) Equilibrium constant $K=5.0$

4) Calculations:

We can construct a RICE table to calculate the changes in the concentrations of products and reactants.

Reaction	$C{H}_{3}H{g\left(glutathione\right)}^{+} \left(aq\right)$	$+cysteine \left(aq\right)$	$\rightleftharpoons C{H}_{3}H{g\left(cysteine\right)}^{+} \left(aq\right)$	$+glutathione \left(aq\right)$
	$\left[C{H}_{3}H{g\left(glutathione\right)}^{+}\right]$ (M)	$\left[cysteine\right]$ (M)	$\left[C{H}_{3}H{g\left(cysteine\right)}^{+} \right]$ (M)	$\left[glutathione \right]$ (M)
Initial	$0.0012$	$0.0012$	$0$	$0$
Change	$- x$	$- x$	$+ x$	$+ x$
Equilibrium	$0.0012-x$	$0.0012-x$	$+x$	$+x$

$Equilibrium constant K_{ }= \frac{\left[C{H}_{3}H{g\left(cysteine\right)}^{+}\right]\left[glutathione\right]}{\left[C{H}_{3}H{g\left(glutathione\right)}^{+}\right]\left[cysteine\right]}$

$K= \frac{x \times x}{\left(0.0012-x\right) \times \left(0.0012-x\right)}$

Since we know the value of equilibrium constant, the change in the concentration can be calculated as follows:

$5= \frac{x^2}{{\left(0.0012-x\right)}^2}$

Calculating the square root on both sides, we get

$\sqrt{5}= \sqrt{\frac{x^2}{{\left(0.0012-x\right)}^2}}$

$2.236= \frac{x^{ }}{{\left(0.0012-x\right)}^{ }}$

$\left(2.236 \right)\times \left(0.0012-x\right)=x$

$0.0026832-2.236x=x$

$0.0026832=x+2.236x=3.236x$

$x= \frac{0.0026832}{3.236}= 8.2917 \times {10}^{-4} M$

From the RICE table, we can see that the equilibrium concentration of glutathione is equal to $x = 8.3\times {10}^{-4} M$.

The equilibrium concentration of cysteine is

$\left(0.0012-x\right)=\left(0.0012-8.2917\times {10}^{-4}\right)= 3.708\times {10}^{-4} M$

Thus, the equilibrium concentration of glutathione is $8.3 \times {10}^{-4}M.$

The equilibrium concentration of cysteine is $3.7 \times {10}^{-4}M.$

Conclusion:

The equilibrium concentration of products and reactants can be calculated using the RICE table for the given equilibrium reaction.