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The pH of a buffer solution containing $0.05 M {Na}_{3} {PO}_{4}$ and $0.1 M {Na}_{2} {HPO}_{4}$ is $11.7$ .The ionization constant ${HPO}_{4}^{2 -}$ is

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$1.0 \times 10^{- 10} M$

$1.0 \times 10^{- 11} M$

$1 \times 10^{- 9} M$

$1.0 \times 10^{- 12} M$

answer is D.

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Detailed Solution

$pH = p K_{a} + \log \frac{[salt]}{[acid]} = p K_{a} + \log (\frac{0.05 M}{0.10 M})$

Hence $11.7 = p K_{a} - 0.30$ .

This give $p K_{a} = 12.0$

Hence, $K_{a} = 1.0 \times 10^{- 12} M$

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