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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2009 · Chemistry · Ionic Equilibrium

Solid Ba (NO_3)_2 is gradually dissolved in a 1.0 × 10^ -4 M Na_2...

Solid Ba (NO3)2\left(\mathrm{NO}_3\right)_2 is gradually dissolved in a 1.0×10−4MNa2CO31.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3 solution. At what concentration of Ba2+\mathrm{Ba}^{2+} will a precipitate begin to form ? \left(\mathrm{K}_{\mathrm{sp}}\right.forBaCO3=5.1×10−9)for \left.\mathrm{Ba} \mathrm{CO}_3=5.1 \times 10^{-9}\right).

Options

  1. A4.1×10−5M4.1 \times 10^{-5} \mathrm{M}
  2. B5.1×10−5M5.1 \times 10^{-5} \mathrm{M}
  3. C8.1×10−8M8.1 \times 10^{-8} \mathrm{M}
  4. D8.1×10−7M8.1 \times 10^{-7} \mathrm{M}

Correct answer

B. 5.1×10−5M5.1 \times 10^{-5} \mathrm{M}

Step-by-step solution

Ba(NO3)2+CaCO3→BaCO3+2NaNO3 Here [CO3−2]=[Na2CO3]=10−4MKsp=[Ba+2][CO3−2]⇒5.1×10−9=[Ba2+](10−4)⇒[Ba+2]=5.1×10−5\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 \times10^{-9}=\left[\mathrm{Ba}^{2+}\right]\left(10^{-4}\right)\Rightarrow \left[\mathrm{Ba}^{+2}\right]=5.1 \times10^{-5} \end{aligned}At this value, just precipitation starts.
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