Ionic EquilibriumHard
Question
H3A is a weak tribasic acid with Ka1 = 10−5, Ka2 = 10−9 and Ka3 = 10−13. The value of pX of 0.1 M – H3A solution, where pX = −log10X and X = $\frac{\left\lbrack A^{3 -} \right\rbrack}{\left\lbrack HA^{2 -} \right\rbrack}$, is
Options
A.5.0
B.4.0
C.9.0
D.10.0
Solution
$\left\lbrack H^{+} \right\rbrack = \sqrt{0.1 \times 10^{- 5}} = 10^{- 3}\text{ M}$
Now, $K_{a_{3}} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{3 -} \right\rbrack}{\left\lbrack HA^{2 -} \right\rbrack} \Rightarrow \frac{\left\lbrack A^{3 -} \right\rbrack}{\left\lbrack HA^{2 -} \right\rbrack} = \frac{10^{- 13}}{10^{- 3}} = 10^{- 10}$
$\therefore P^{x} = 10$
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