Chapter 22, Problem 22.59QA

Interpretation Introduction

To:

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

Answer to Problem 22.59QA

Solution:

$pH=3.06$

Explanation of Solution

1) Concept:

We are asked to find the pH of a solution of selenocysteine, which is a diprotic acid with two $p{K}_a$ values as 2.21 and 5.43 respectively. Hydronium ion concentration of selenocysteine can be calculated using the RICE table for dissociation of each of the two ionizable protons in selenocysteine and pH is calculated. We would use a notation $H_{2}A$ for selenocysteine, as it is a diprotic acid.

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) $\left[selenocysteine\right]=\left[H_{2}A\right]=1.00\times {10}^{-3}M$      

ii) $p{K}_{a1}=2.21$

iii)  $p{K}_{a2}=5.43$

4) Calculations:

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

Reaction	$H_{2}A\left(aq\right) + { H}_{2}O \left(l\right) \rightleftharpoons H{A}^{-}\left(aq\right) + H_3{O}^{+}\left(aq\right)$
	$\left[H_{2}A\right] \left(M\right)$		[$H{A}^{-} \left] \right(M)$	[$H_3{O}^{+}\left] \right(M)$
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 above reaction is

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

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

${10}^{-2.21}=\frac{x^2}{(1.00\times {10}^{-3}-x)}$

$6.16595\times {10}^{-3}= \frac{x^2}{\left(1.00\times {10}^{-3}-x\right)}$

$[\left(6.16595\times {10}^{-3}\right)\times \left(1.00\times {10}^{-3}-x\right)]=x^2$

$6.16595\times {10}^{-6}-6.16595\times {10}^{-3}x=x^2$

$x^2+ 6.16595\times {10}^{-3}x- 6.16595\times {10}^{-6}=0\dots \left(1\right)$

This equation fits the general form of a quadratic equation. Solving this equation to get values of x as

$x=8.75647\times {10}^{-4}$

$x=\mathrm{ }-7.0416\times {10}^{-3}$

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

$\left[H_3{O}^{+}\right]=x=8.75647\times {10}^{-4}\mathrm{ }M=\left[H{A}^{-}\right]$

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

RICE table for selenocysteine [HA] for 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	$8.75647\times {10}^{-4}\mathrm{ }$		$0$	$8.75647\times {10}^{-4}\mathrm{ }$
Change	$-x$		$+x$	$+x$
Equilibrium	$(8.75647\times {10}^{-4}\mathrm{ }-x)$		$x$	$(8.75647\times {10}^{-4}\mathrm{ }+x$)

Equilibrium constant expression for the second dissociation for selenocysteine is

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

${10}^{-5.43}=\frac{\left(x\right) (8.75647\times {10}^{-4}\mathrm{ }+x)}{(8.75647\times {10}^{-4}\mathrm{ }-x)}$

$3.71535 \times {10}^{-6}= \frac{\left(x\right) (8.75647\times {10}^{-4}\mathrm{ }+x)}{(8.75647\times {10}^{-4}\mathrm{ }-x)}$

$(3.71535 \times {10}^{-6})(8.75647\times {10}^{-4}-x)= \left(x\right) (8.75647\times {10}^{-4}+x)$

$3.253335081\times {10}^{-9}-3.71535 \times {10}^{-6}x= 8.75647\times {10}^{-4}x+x^2$

$x^2+ 8.7936235\times {10}^{-4}x-3.253335081 \times {10}^{-9}=0$

Solving this quadratic equation to get two roots as

$x=3.683132179\times {10}^{-6}$ And  $x=-8.833\times {10}^{-4}$

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

Thus, $\left[H_3{O}^{+}\right]=(3.683132179\times {10}^{-6}+ 8.75647\times {10}^{-4}) =8.79330132179\times {10}^{-4}$

Calculating pH using the formula

$pH= -\log \left[H_3{O}^{+}\right]= -\log (8.79330132179\times {10}^{-4} )$

$pH= -\mathrm{l}\mathrm{o}\mathrm{g}(8.79330132179\times {10}^{-4})$

$pH=3.06$

Conclusion:

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