An aqueous solution contains an unknown concentration of $\mathrm{Ba}^{2+}$. When 50 mL of a 1 M solution of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is added, $\mathrm{BaSO}_{4}$ just begins to precipitate. The final volume is 500 mL . The solubility product of $\mathrm{BaSO}_{4}$ is $1 \times 10^{-10}$. What is the original concentration of $\mathrm{Ba}^{2+}$ ?

(D) $1.1 \times 10^{-9} \mathrm{M}$

(A) $1.0 \times 10^{-10} \mathrm{M}$

(B) $5 \times 10^{-9} \mathrm{M}$

(C) $2 \times 10^{-9} \mathrm{M}$

JEE Main · 2018 · 08 Apr 2018 · Offline · Equilibrium

An aqueous solution contains an unknown concentration of

Ba^{2 +}

. When 50 mL of a 1 M solution of

Na_{2} SO_{4}

is added,

BaSO_{4}

just begins to precipitate. The final volume is 500 mL . The solubility product of

BaSO_{4}

1 \times 1 0^{- 10}

. What is the original concentration of

Ba^{2 +}

Answer: (D)

Precipitation just begins when the ion product reaches

K_{s p}

, so

[Ba X^{2 +}] [SO X_{4} X^{2 -}] = 1 \times 1 0^{- 10}

. The sulfate added,

1

M in

50

mL diluted to the final

500

mL, gives

[SO X_{4} X^{2 -}] = \frac{1 \times 50}{500} = 0.1

M, so at the onset

[Ba X^{2 +}] = \frac{1 0 ^{- 10}}{0.1} = 1 0^{- 9}

M in that

500

mL. The barium was originally dissolved in only

450

mL, before the

50

mL was added, so its starting concentration was

\frac{1 0 ^{- 9} \times 500}{450} = 1.1 \times 1 0^{- 9}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option