Chapter 22, Problem 22.60QA

Interpretation Introduction

To:

a)  Calculate the $pH$ of $1.00\times {10}^{-3}M$ solution of cysteine.

b)  Explain if selenocysteine is a stronger acid than cysteine.

Answer to Problem 22.60QA

Solution:

a)  $pH=3.02$

b) No, selenocysteine is not a stronger acid than cysteine.

Explanation of Solution

1) Concept:

We are asked to find the pH of a solution of cysteine, which is a diprotic acid with two $p{K}_a$ values as 1.7 and 8.3 respectively. Hydronium ion concentration of cysteine can be calculated using the RICE table for dissociation of each of the two ionizable protons in cysteine and pH is calculated. We would use a notation $H_{2}A$ for cysteine, as it is a diprotic acid. Smaller the $p{K}_{a1}$ value stronger is the acid. The $p{K}_{a1}$ for cysteine is 1.7 which is smaller than that of selenocysteine which is 2.21. Hence, selenocysteine is not a stronger acid than cysteine.

2) Formula:

i) $p{K}_a=-\log \left[K_a\right]$

ii) $pH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[H_3{O}^{+}\right]$

3) Given:

i) $\left[cysteine\right]=1.00\times {10}^{-3}M$      

ii) $p{K}_{a1}=1.7$

iii)  $p{K}_{a1}=8.3$

4) Calculations:

RICE table for cysteine $\left[H_{2}A\right]$ for the first dissociation:

Reaction	$H_{2}A\left(aq\right) + { H}_{2}O \left(l\right) \rightleftharpoons H{A}^{-}\left(aq\right) + H_3{O}^{+}\left(aq\right)$
	$\left[H_{2}A\right] \left(M\right)$		$\left[H{A}^{-}\right] \left(M\right)$	[$H_3{O}^{+}\left] \right(M)$
Initial	$1.00\times {10}^{-3}$		$0$	$0$
Change	$-x$		$+x$	$+x$
Equilibrium	$(1.00\times {10}^{-3}-x)$		$x$	$x$

Equilibrium constant expression for the above reaction is

$K_{a1}=\frac{\left[H{A}^{-} \right]\left[H_3{O}^{+}\right] }{\left[H_{2}A\right]}$

$K_{a1}=\frac{x^2}{(1.00\times {10}^{-3}-x)}$

${10}^{-1.7}=\frac{x^2}{(1.00\times {10}^{-3}-x)}$

$1.995262\times {10}^{-2}= \frac{x^2}{\left(1.00\times {10}^{-3}-x\right)}$

$\left(1.995262\times {10}^{-2}\right)\times \left(1.00\times {10}^{-3}-x\right)=x^2$

$1.995262\times {10}^{-5}-1.995262\times {10}^{-2}x=x^2$

$x^2+ 1.995262\times {10}^{-2}x- 1.995262\times {10}^{-5}=0$

This equation fits the general form of a quadratic equation, solving this equation to get values of $x\mathrm{ }$ as:

$x=9.5435\times {10}^{-4}$

$x=\mathrm{ }-2.09\times {10}^{-2}$

We will use positive value of $x$ for further calculation, because the concentration can never be negative.

$\left[H_3{O}^{+}\right]=x=9.5435\times {10}^{-4}M=\left[H{A}^{-}\right]$

We would use these equilibrium concentrations for $H{A}^{-}$ and $H_3{O}^{+}$ as initial concentration for the second dissociation of cysteine.

RICE table for selenocysteine for the second dissociation:

Reaction	$H{A}^{-}\left(aq\right) + { H}_{2}O \left(l\right) \rightleftharpoons A^{2-}\left(aq\right) + H_3{O}^{+}\left(aq\right)$
	$\left[H{A}^{-}\right] \left(M\right)$		[$A^{2-} \left] \right(M)$	[$H_3{O}^{+}\left] \right(M)$
Initial	$9.5435\times {10}^{-4}\mathrm{ }$		$0$	$9.5435\times {10}^{-4}\mathrm{ }$
Change	$-x$		$+x$	$+x$
Equilibrium	$(9.5435\times {10}^{-4}\mathrm{ }-x)$		$x$	$(9.5435\times {10}^{-4}\mathrm{ }+x$)

Equilibrium constant expression for the second dissociation for selenocysteine is

$K_{a2}=\frac{\left[A^{2-} \right]\left[H_3{O}^{+}\right] }{\left[H{A}^{-}\right]}$

${10}^{-8.3}=\frac{\left(x\right) (9.5435\times {10}^{-4}\mathrm{ }+x)}{(9.5435\times {10}^{-4}\mathrm{ }-x)}$

$5.01187 \times {10}^{-9}= \frac{\left(x\right) (9.5435\times {10}^{-4}\mathrm{ }+x)}{(9.5435\times {10}^{-4}\mathrm{ }-x)}$

$(5.01187 \times {10}^{-9})(9.5435\times {10}^{-4}-x)= \left(x\right) (9.5435\times {10}^{-4}+x)$

$4.783080\times {10}^{-12}-5.01187 \times {10}^{-9}x= 9.5435\times {10}^{-4}x+x^2$

$x^2+\left( 9.5435\times {10}^{-4}+5.01187 \times {10}^{-9}\right)x-4.78307\times {10}^{-12}=0$

$x^2+9.5435\times {10}^{-4}x-4.78307\times {10}^{-12}=0$

Solving this quadratic equation to get two roots as

$x=5.01183\times {10}^{-9}$ and  $x=-9.5435\times {10}^{-4}$

We will consider only the positive value of $x$, since concentrations can never be negative.

Thus, $\left[H_3{O}^{+}\right]=\left(9.5435\times {10}^{-4}+5.01183\times {10}^{-9}\right)=9.54355\times {10}^{-4}M$

Calculating pH using the formula:

$pH= -\log \left[H_3{O}^{+}\right]$

$pH=-\log \left(9.54355\times {10}^{-4}\right)$

$pH=3.02$ 0

Therefore, the $pH$ of $1.00\times {10}^{-3}M$ cysteine solution is $3.02$.

Conclusion:

For a diprotic acid, we need to consider the dissociation of two protons to find the pH of  the solution.