Concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in a solution is 1 M and \(1.8 \times 10^{-2} \mathrm{M}\), respectively. Molar solubility of \(\mathrm{PbSO}_{4}\) in the same solution is \(\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}\) (expressed in scientific notation). The value of Y is ____. [Given: Solubility product of \(\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}\). For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), Ka1 is very large and \(K_{a 2}=1.2 \times 10^{-2}\)]

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Equilibrium

Concentration of

H_{2} SO_{4}

and

Na_{2} SO_{4}

in a solution is 1 M and

1.8 \times 1 0^{- 2} M

, respectively. Molar solubility of

PbSO_{4}

in the same solution is

X \times 1 0^{- Y} M

(expressed in scientific notation). The value of Y is ____.
[Given: Solubility product of

PbSO_{4} (K_{s p}) = 1.6 \times 1 0^{- 8}

. For

H_{2} SO_{4}

, Ka1 is very large and

K_{a 2} = 1.2 \times 1 0^{- 2}

]

Answer: 6

First, determine the concentration of

SO_{4}^{2 -}

ions in solution.

H_{2} SO_{4}

fully dissociates in the first step to give 1 M

HSO_{4}^{-}

and 1 M

H^{+}

Na_{2} SO_{4}

contributes

1.8 \times 1 0^{- 2}

SO_{4}^{2 -}

. For the second dissociation of

HSO_{4}^{-}

, using

K_{a 2} = 1.2 \times 1 0^{- 2}

and the equilibrium expression, we find the additional

SO_{4}^{2 -}

from

HSO_{4}^{-}

dissociation. Since

Q_{C} > K_{C}

, the equilibrium shifts backward. Solving gives

[SO_{4}^{2 -}] = 1.2 \times 1 0^{- 2}

M total.

For

PbSO_{4}

solubility:

K_{s p} = s (s + 1.2 \times 1 0^{- 2}) = 1.6 \times 1 0^{- 8}

. Since

s ≪ 1.2 \times 1 0^{- 2}

, we have

s \times 1.2 \times 1 0^{- 2} = 1.6 \times 1 0^{- 8}

, giving

s = \frac{1.6}{1.2} \times 1 0^{- 6} = 1.33 \times 1 0^{- 6}

M. Thus

Y = 6

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer