Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

Solid Ba (NO3)2Ba\, (NO_3)_2 is gradually dissolved in a 1.0?10−4M Na2CO31.0 ? 10^{-4}M\,Na_2CO_3 solution. At what concentration of Ba2+Ba^{2+} will a precipitate begin to form ?(KspK_{sp} for Ba CO3=5.1?10−9Ba\, CO_3 = 5.1 ? 10^{-9}).

A

4.1?10−5M4.1 ?10^{-5}M

B

5.1?10−5M5.1 ?10^{-5}M

C

8.1?10−8M8.1 ?10^{-8}M

D

8.1?10−7M8.1 ?10^{-7}M

Answer

5.1?10−5M5.1 ?10^{-5}M

Explanation

Solution

Ba(NO3)2+CaCO3→BaCO3+2NaNO3Ba\left(NO_{3}\right)_{2}+CaCO_{3} \rightarrow BaCO_{3}+2NaNO_{3} Here [CO3−2]=[Na2CO3]=10−4M\left[CO^{-2}_{3}\right]=\left[Na_{2}CO_{3}\right]=10^{-4}M Ksp=[Ba+2][CO3−2]⇒5.1×10−9=[Ba2+][10−4]⇒[Ba2+]=5.1×10−5K_{sp}=\left[Ba^{+2}\right]\left[CO^{-2}_{3}\right] \Rightarrow 5.1\times10^{-9}=\left[Ba^{2+}\right]\left[10^{-4}\right] \Rightarrow \left[Ba^{2+}\right]=5.1\times10^{-5} At this value, just precipitation starts.