Solid $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is gradually dissolved in a $1.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ solution. At what concentration of $\mathrm{Ba}^{2+}$ will a precipitate begin to form? $\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\left.\mathrm{Ba} \mathrm{CO}_{3}=5.1 \times 10^{-9}\right)$.

(B) $5.1 \times 10^{-5} \mathrm{M}$

(A) $4.1 \times 10^{-5} \mathrm{M}$

(C) $8.1 \times 10^{-8} \mathrm{M}$

(D) $8.1 \times 10^{-7} \mathrm{M}$

JEE Main · 2009 · 12 Apr 2009 · Offline · Equilibrium

Solid

Ba (NO_{3})_{2}

is gradually dissolved in a

1.0 \times 1 0^{- 4} M Na_{2} CO_{3}

solution. At what concentration of

Ba^{2 +}

will a precipitate begin to form?

(K_{sp}

for

Ba CO_{3} = 5.1 \times 1 0^{- 9})

Answer: (B)

A precipitate of barium carbonate first appears when the ionic product just reaches the solubility product,

K_{s p} = [B a^{2 +}] [C O_{3}^{2 -}]

. The carbonate concentration is fixed by the sodium carbonate at

1.0 \times 1 0^{- 4}

M, so the barium concentration needed is

[B a^{2 +}] = \frac{5.1 \times 1 0 ^{- 9}}{1.0 \times 1 0 ^{- 4}}

, which is

5.1 \times 1 0^{- 5}

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option