Solid Ba (NO_3 )_2 is gradually dissolved in a 1.0 × 10^-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 × 10^-9 ) .

Correct answer: (B) 5.1 × 10^-5 M. Ba (NO_3 )_2+CaCO_3 arrow BaCO_3+2 NaNO_3 Here [CO_3^-2 ]= [Na_2 CO_3 ]=10^-4 M K_sp= [Ba^+2 ] [CO_3^-2 ] ⇒ 5.1 × 10^-9= [Ba^2+ ] (10^-4 ) ⇒ [Ba^+2 ]=5.1 × 10^-5 At this value, just precipitation starts.

Q17 / 90 · JEE Main 2009 · Chemistry

Solid 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 $\left.\mathrm{Ba} \mathrm{CO}_3=5.1 \times 10^{-9}\right)$.

A
$4.1 \times 10^{-5} \mathrm{M}$
Correct answer
B
$5.1 \times 10^{-5} \mathrm{M}$
C
$8.1 \times 10^{-8} \mathrm{M}$
D
$8.1 \times 10^{-7} \mathrm{M}$

Solution

$$ \begin{aligned} & \mathrm{Ba}\left(\mathrm{NO}_3\right)_2+\mathrm{CaCO}_3 \rightarrow \mathrm{BaCO}_3+2 \mathrm{NaNO}_3 \\ & \text { Here }\left[\mathrm{CO}_3^{-2}\right]=\left[\mathrm{Na}_2 \mathrm{CO}_3\right]=10^{-4} \mathrm{M} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{CO}_3^{-2}\right] \Rightarrow 5.1 \times 10^{-9}=\left[\mathrm{Ba}^{2+}\right]\left(10^{-4}\right) \Rightarrow\left[\mathrm{Ba}^{+2}\right]=5.1 \times 10^{-5} \end{aligned} $$ At this value, just precipitation starts.

Hence, the correct answer is option: 'B'.

Solution

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