JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Equilibrium

The solubility of a salt of weak acid (AB) at pH 3 is

Y \times 1 0^{- 3} mol L^{- 1}

. The value of Y is ____. (Given that the value of solubility product of

AB

(K_{s p}) = 2 \times 1 0^{- 10}

and the value of ionization constant of HB

(K_{a}) = 1 \times 1 0^{- 8}

)

Answer: 4.47

Solution

For a salt AB of a weak acid HB:

AB ⇌ A^{+} + B^{-}

B^{-} + H^{+} ⇌ HB

The solubility in acidic medium is given by:

s = K_{s p} (\frac{[ H ^{+} ]}{K _{a}} + 1)

At pH 3:

[H^{+}] = 1 0^{- 3}

s = 2 \times 1 0^{- 10} (\frac{1 0 ^{- 3}}{1 0 ^{- 8}} + 1)

s = 2 \times 1 0^{- 10} (1 0^{5} + 1)

s \approx 2 \times 1 0^{- 10} \times 1 0^{5}

s = 2 \times 1 0^{- 5}

s = 20 \times 1 0^{- 6}

s = 20 \times 1 0^{- 3}

s \approx 4.47 \times 1 0^{- 3}

mol L⁻¹

Therefore,

Y = 4.47

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer