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 $0.1M{H}_{3}A(aq.)$ solution, where $pX=-\log X$ 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 $K_{a1}$
$K_{a1}=\frac{\left[H_2{A}^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_{3}A\right]}$
${10}^{-5}=\frac{Y\times Y}{0.1-Y}$
Since Ka1 is small, we can approximate 0.1-Y to 0.1.
${10}^{-5}=\frac{Y\times Y}{0.1}$
${10}^{-5}\times 0.1=Y^2$
$Y={10}^{-3}=\left[H_3{O}^{+}\right]$
Now from $K{a}_3$, we will calculate X
$X=\frac{\left[A^{3-}\right]}{\left[H{A}^{2-}\right]}$
$K_{a_3}=\frac{\left[A^{3-}\right]\left[H_3{O}^{+}\right]}{\left[H{A}^{2-}\right]}$
$K_{a_3}=X\left[H_3{O}^{+}\right]$
${10}^{-13}=X\times {10}^{-3}$
$X={10}^{-10}$
$pX=-logX=-log{10}^{-10}=10$