Chapter 23, Problem 23.83QA

Interpretation Introduction

To find:

The equilibrium concentration of cysteine and penicillamine.

Answer to Problem 23.83QA

Solution:

[Penicillamine] = $9.98\times {10}^{-4} M$

[Cysteine] = $1.00 M$

Explanation of Solution

1) Concept:

The equilibrium reaction is:

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

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

2) Formula:

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

3) Given:

i) Initial concentration of cysteine $=1.0 M$

ii) Initial concentration of $C{H}_{3}H{g\left(penicillamine\right)}^{+}=1.0 mM= 1.0 \times {10}^{-3} M$

iii) Equilibrium constant $K=0.633$

4) Calculations:

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

Reaction	$C{H}_{3}H{g\left(penicillamine\right)}^{+} \left(aq\right)$	$+cysteine \left(aq\right)$	$\rightleftharpoons C{H}_{3}H{g\left(cysteine\right)}^{+} \left(aq\right)$	$+penicillamine \left(aq\right)$
	$\left[C{H}_{3}H{g\left(penicillamine\right)}^{+} \right]$ (M)	$\left[cysteine\right]$ (M)	$\left[C{H}_{3}H{g\left(cysteine\right)}^{+}\right]$ (M)	$\left[penicillamine\right]$ (M)
Initial	$1.0 \times {10}^{-3}$	$1.0$	$0$	$0$
Change	$- x$	$- x$	$+ x$	$+ x$
Equilibrium	$1.0 \times {10}^{-3}-x$	$1.0-x$	$+x$	$+x$

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

$K= \frac{x \times x}{\left(1.0 \times {10}^{-3}-x\right) \times \left(1.0-x\right)}$

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

$0.633= \frac{x^2}{\left(1.0 \times {10}^{-3}-x\right) \times \left(1.0-x\right)}$

$0.633 \times \left(1.0 \times {10}^{-3}-x\right) \times \left(1.0-x\right)=x^2$

$0.633 \times \left(0.001-1.001x+x^2\right)=x^2$

$0.000633-0.633633x+0.633x^2=x^2$

$0.367x^2+0.633633x-0.000633=0$

On solving the above quadratic equation, the value of  $x$ is $9.984 \times {10}^{-4} M$.

From the RICE table, we can see that the equilibrium concentration of penicillamine is equal to $x=9.98\times {10}^{-4} M.$

The equilibrium concentration of cysteine is

 $\left(1.0 -x\right)M=\left(1.0- 9.9984 \times {10}^{-4}\right)M=0.999 M$

Thus, the equilibrium concentration of penicillamine is $9.98 \times {10}^{-4}M.$

The equilibrium concentration of cysteine is $1.00 M$.

Conclusion:

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