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JEE Main levelSingle correct optionDifficulty 2 of 3ia007

The degree of hydrolysis of 0.010 M sodium fluoride at 298 K is closest to ( $K_{a}$ of HF $= 6.8 \times 1 0^{- 4}$ ; use $h = K_{h} / c$ )

A $3.8 \times 1 0^{- 7}$
B $3.8 \times 1 0^{- 5}$
C $1.5 \times 1 0^{- 9}$
D $2.6 \times 1 0^{- 1}$

Answer: B

K_{h} = K_{w} / K_{a} = 1.47 e - 11

. Then

h = K_{h} / c = 1.47 e - 11/0.010 = 3.83 e - 05

. The wrong choices use

K_{h} c

(which is

[OH X^{-}]

),

K_{h} / c

without the root, or

K_{a}

in place of

K_{h}

.

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