Question
A volume of 10 ml of 0.1 M tribasic acid, H3A is titrated with 0.1 M-NaOH solution. What is the ratio (approximate value) of $\frac{\left\lbrack H_{3}A \right\rbrack}{\left\lbrack A^{3 -} \right\rbrack}$ at the second equivalent point? Given: K1 = 7.5 × 10−4; K2 = 10−8; K3 = 10−12
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Solution
At 2nd equation point: $P^{H} = \frac{1}{2}\left( P^{K_{a2}} + P^{K_{a3}} \right) = \frac{8 + 12}{12} = 10$
Now, $K_{a1}.K_{a2}.K_{a3} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{3 -} \right\rbrack}{\left\lbrack H_{3}A \right\rbrack}$
Or, $7.5 \times 10^{- 4} \times 10^{- 8} \times 10^{- 12} = \frac{\left( 10^{- 10} \right)^{3} \times \left\lbrack A^{3 -} \right\rbrack}{\left\lbrack H_{3}A \right\rbrack}$
$\therefore\frac{\left\lbrack H_{3}A \right\rbrack}{\left\lbrack A^{3 -} \right\rbrack} = \frac{10^{- 6}}{7.5} = 1.33 \times 10^{- 7}$
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