Question

$H_{3}A$ is a weak triprotic acid $(K{a}_1={10}^{-5},K{a}_2={10}^{-9},K{a}_3={10}^{-13}).$
What is the value of $pX$ of $0.1M{H}_{3}A\left(aq\right)$ solution?
Given $pX=-logX$ and $X=\frac{\left[A^{3-}\right]}{\left[H{A}^{2-}\right]}$.

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The correct option is D 10

First we will calculate $\left[H_3{O}^{+}\right]$ from $K{a}_1$
At equilibrium :

$H_{3}A(aq.)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}(aq.)+H_2{A}^{-}(aq.)0.1-YYY$


$K{a}_1=\frac{\left[H_2{A}^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_{3}A\right]}$

${10}^{-5}=\frac{Y\times Y}{0.1-Y}$
since $K{a}_1$ is small , $0.1-Y\approx 0.1$

${10}^{-5}=\frac{Y\times Y}{0.1}$

${10}^{-6}=Y\times Y$

$\Rightarrow Y^2={10}^{-6}$

$\Rightarrow Y={10}^{-3}=\left[H_3{O}^{+}\right]$


Now from $K{a}_3$, we will calculate $X$
where,
$X=\frac{\left[A^{3-}\right]}{\left[H{A}^{2-}\right]}$
At equilibrium,
$H{A}^{2-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+A^{3-}\left(aq\right)$

$K{a}_3=\frac{\left[A^{3-}\right]\left[H_3{O}^{+}\right]}{\left[H{A}^{2-}\right]}$

$\Rightarrow K{a}_3=X\times \left[H_3{O}^{+}\right]$

putting value of $\left[H_3{O}^{+}\right]$ and $K{a}_3$,

$\Rightarrow {10}^{-13}=X\times {10}^{-3}$

$\Rightarrow X={10}^{-10}$

$\Rightarrow pX=-\text{log}X=-\text{log}\left({10}^{-10}\right)=10$



Theory :
Polyprotic acids :
Acids which supply/provide more than one $H^{+}$ per molecule.
Consider dissociation of a triprotic acid $H_{3}P{O}_4$ :
First step :
$H_{3}P{O}_4\rightleftharpoons H^{+}\left(aq\right)+H_{2}P{O}_4^{-}\left(aq\right)$
$C-x(x+y+z)(x-y)$
Second Step :
$H_{2}P{O}_4^{-}\rightleftharpoons H^{+}\left(aq\right)+HP{O}_4^{2-}\left(aq\right)$
$(x-y)(x+y+z)(y-z)$
Third step :
$HP{O}_4^{2-}\rightleftharpoons H^{+}\left(aq\right)+P{O}_4^{3-}\left(aq\right)$
$(y-z)(x+y+z)\left(z\right)$
Dissociation constant for First step :
$K_{a_1}=\frac{\left[H^{+}\right]\left[H_{2}P{O}_4^{-}\right]}{\left[H_{3}P{O}_4\right]}$
Dissociation constant for second step :
$K_{a_2}=\frac{\left[H^{+}\right]\left[HP{O}_4^{2-}\right]}{\left[H_{2}P{O}_4^{-}\right]}$
Dissociation constant for third step :
$K_{a_3}=\frac{\left[H^{+}\right]\left[P{O}_4^{3-}\right]}{\left[HP{O}_4^{2-}\right]}$
Therefore :
Dissociation constant for first step :
$K_{a_1}=\frac{(x+y+z)(x-y)}{C-x}$
Approximations for first,second and third step :
$x-y\approx x$,
$x+y+z\approx x$ and $y-z\approx y$
Therefore Dissociation constant for all the three steps can also be written as :
$K_{a_1}=\frac{\left(x\right)\left(x\right)}{C_1-x}$

$K_{a_2}=\frac{\left(x\right)\left(y\right)}{\left(x\right)}=y$

$K_{a_3}=\frac{\left(x\right)\left(z\right)}{\left(y\right)}$