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Chemistry Question on Acids and Bases

Solid Ba(NO3)2Ba(NO_3)_2 is gradually dissolved in a 1.0×10−41.0 \times 10^{-4} M Na2CO3Na_2CO_3 solution. At which concentration of Ba2+Ba^{2+}, precipitate of BaCO3BaCO_3 begins to form ? (KspK_{sp} for BaCO3=5.1×10−9BaCO_3 = 5.1 \times 10^{-9})

A

5.1×10−5 M5.1 \times 10^{-5}\, M

B

7.1×10−8 M7.1 \times 10^{-8}\,M

C

4.1×10−5 M4.1 \times 10^{-5}\, M

D

8.1×10−7 M8.1 \times 10^{-7}\,M

Answer

5.1×10−5 M5.1 \times 10^{-5}\, M

Explanation

Solution

Given Na2CO3=1.0×10−4M\quad Na_{2}CO_{3} = 1.0 \times 10^{-4}M
∴[CO3−−]=1.0×10−4M\therefore \left[CO_{3} ^{--}\right] = 1.0 \times 10^{-4} M
i.e.s=1.0×10−4Mi.e.\quad s=1.0\times10^{-4}M
At equilibrium
[Ba++][CO3−−]=Ksp\left[Ba^{++}\right] \left[CO_{3}^{--}\right] = K_{sp} of BaCO3BaCO_{3}
[Ba++]=Ksp[CO3−−]=5.1×10−91.0×10−4\left[Ba^{++}\right] = \frac{K_{sp}}{ \left[CO_{3}^{--}\right]} = \frac{5.1\times10^{-9}}{1.0\times10^{-4}}
=5.1×10−5M= 5.1\times 10^{-5} M