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JEE Main levelSingle correct optionDifficulty 2 of 3ia181

A diprotic acid $H X_{2} A$ has $p K_{a 1} = 4.0$ and $p K_{a 2} = 8.0$ . The fraction of the total acid present as $HA X^{-}$ at $pH = 5.0$ is closest to

A $0.91$
B $0.09$
C $0.99$
D $0.50$

Answer: A

[H X_{2} A] / [HA X^{-}] = 1 0^{- 5} /1 0^{- 4} = 0.1

and

[A X^{2 -}] / [HA X^{-}] = 1 0^{- 8} /1 0^{- 5} = 1 0^{- 3}

. Fraction of

HA X^{-} = 1/ (1 + 0.1 + 0.001) = 0.908

.

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