Chapter 21, Problem 21.60QP

Interpretation Introduction

Interpretation: The $\text{pH}$ value of $1.00\times {10}^{-3}\text{M}$ solution of cysteine is to be calculated. Selenocysteine being a stronger acid than cysteine is to be validated.

Concept introduction: The negative logarithm of the hydrogen ion concentration of a solution is known as $\text{pH}$.

The $\text{pH}$ is calculated by the formula,

$\text{pH=}-\log \left[{\text{H}}^{+}\right]$

To determine: The $\text{pH}$ value of $1.00\times {10}^{-3}\text{M}$ solution of cysteine and if selenocysteine is a stronger acid than cysteine.

Answer to Problem 21.60QP

Solution

The $\text{pH}$ value of $1.00\times {10}^{-3}\text{M}$ solution of cysteine is $\underset{\_}{4.80}$.

Selenocysteine is a stronger acid than cysteine.

Explanation of Solution

Explanation

Given

The concentration of solution of cysteine is $1.00\times {10}^{-3}\text{M}$.

The value of $\text{p}{K}_{{\text{a}}_{\text{1}}}$ is $1.7$.

The value of $\text{p}{K}_{{\text{a}}_2}$ is $8.3$.

The negative logarithm of the hydrogen ion concentration of a solution is known as $\text{pH}$.

The $\text{pH}$ is calculated by the formula,

$\text{pH=}-\log \left[{\text{H}}^{+}\right]$ (1)

Where,

• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ ions.

The negative logarithm to the equilibrium constant for the acidic reaction is known as $\text{p}{K}_{\text{a}}$. It is calculated by the formula,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Where,

• $K_{\text{a}}$ is equilibrium constant for the acidic reaction.

The value of $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}={10}^{-\text{p}{K}_{\text{a}}}$ (2)

The reaction of selenocysteine in ionized form is,

$\text{Cysteine}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{cysteine}}^{-}\left(aq\right)$

The equilibrium constant for the above reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{cysteine}}^{-}\right]}{\left[\text{Cysteine}\right]}$ (3)

Where,

• $K_{\text{a}}$ is equilibrium constant for the acidic reaction.
• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ ions.
• $\left[{\text{cysteine}}^{-}\right]$ is concentration of cysteine ions.
• $\left[\text{Cysteine}\right]$ is concentration of cysteine.

The value of $\text{p}{K}_{\text{a}}$ is calculated by the formula,

$\text{p}{K}_{\text{a}}=\text{p}{K}_{a_2}-\text{p}{K}_{a_1}$

Substitute the given values of $\text{p}{K}_{{\text{a}}_{\text{1}}}$ and $\text{p}{K}_{{\text{a}}_2}$ in the above expression.

$\begin{array}{c}\text{p}{K}_{\text{a}}=8.3-1.7\\ =6.6\end{array}$

Substitute the value of $\text{p}{K}_{\text{a}}$ in equation $\left(2\right)$ to get the value of $K_{\text{a}}$.

$\begin{array}{c}{K}_{\text{a}}={10}^{-6.6}\\ =2.51\times {10}^{-7}\end{array}$

In the beginning of the reaction,

The initial concentration of cysteine is $1.00\times {10}^{-3}\text{M}$.

The initial concentration of cysteine ions is $0$.

The initial concentration of ${\text{H}}^{+}$ ions is $0$.

At equilibrium,

The moles of reactant (cysteine) that is used up is assumed to be $x$.

Therefore, concentration of cysteine is,

$\left[\text{Cysteine}\right]=\left(1.00\times {10}^{-3}-x\right)\text{M}$

The value of $x$ is very small in comparison with $1.00\times {10}^{-3}\text{M}$. So,

$\begin{array}{c}\left[\text{Cysteine}\right]=\left(1.00\times {10}^{-3}-x\right)\text{M}\\ \approx 1.00\times {10}^{-3}\text{M}\end{array}$

The concentration of product is assumed to be $x$ that is increased.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\left[{\text{Cysteine}}^{-}\right]\\ =x\text{M}\end{array}$

The table to show concentration of reactant and products

$\begin{array}{cccc}& \left[\text{Cysteine}\right]\text{M}& \left[{\text{Cysteine}}^{-}\right]\text{M}& \left[{\text{H}}^{+}\right]\text{M}\\ \text{Intial}& 1.00\times {10}^{-3}& 0& 0\\ \text{Change}& -x& +x& +x\\ \text{Equlibrium}& 1.00\times {10}^{-3}-x\cong 1.00\times {10}^{-3}& x& x\end{array}$

Substitute the values of $K_{\text{a}}$, $\left[\text{Cysteine}\right]$, $\left[{\text{Cysteine}}^{-}\right]$ and $\left[{\text{H}}^{+}\right]$ in the equation $\left(3\right)$

$\begin{array}{c}2.51\times {10}^{-7}=\frac{\left(x\right)\left(x\right)}{1.00\times {10}^{-3}\text{M}}\\ x=1.58\times {10}^{-5}\text{M}\end{array}$

Therefore, concentration of ${\text{H}}^{+}$ is $1.58\times {10}^{-5}\text{M}$.

Substitute the value of concentration of ${\text{H}}^{+}$ in equation $\left(1\right)$ to calculate the value of $\text{pH}$.

$\begin{array}{c}\text{pH}=-\text{log}\left(1.58\times {10}^{-5}\right)\\ =\underset{\_}{4.80}\end{array}$

Therefore, $\text{pH}$ value of $1.00\times {10}^{-3}\text{M}$ solution of cysteine is $\underset{\_}{4.80}$.

The value of $\text{p}{K}_{\text{a}}$ determines the acidity. If $\text{p}{K}_{\text{a}}$ is greater, then the acidity is lower and vice versa.

The $\text{p}{K}_{\text{a}}$ value of selenocysteine is $3.22$ and the $\text{p}{K}_{\text{a}}$ value of cysteine is $6.6$.

The $\text{p}{K}_{\text{a}}$ value of selenocysteine is less than the $\text{p}{K}_{\text{a}}$ value of cysteine.

Therefore, selenocysteine is a stronger acid than cysteine.

Conclusion

The $\text{pH}$ value of $1.00\times {10}^{-3}\text{M}$ solution of cysteine is $\underset{\_}{4.80}$.

Selenocysteine is a stronger acid than cysteine.