Equilibrium Question 807
Question: Solid $ Ba{{(NO_3)} _2} $ is gradually dissolved in a $ 1.0\times {10^{-4}}M,Na _2CO _3 $ solution. At which concentration of $ B{a^{2+}} $ precipitate of $ BaCO _3 $ begins to form- ( $ K _{sp} $ for $ BaCO _3=5.1\times {10^{-9}} $ )
Options:
A) $ 5.1\times {10^{-5}}M $
B) $ 7.1\times {10^{-8}}M $
C) $ 4.1\times {10^{-5}}M $
D) $ 8.1\times {10^{-7}}M $
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ Na_2CO_3=1.0\times {10^{-4}}M $
$ \therefore [ CO_3^{2-} ]=1.0\times {10^{-4}}M $ i.e. $ s=1.0\times {10^{-4}}M $ At equilibrium $ [B{a^{2+}}][CO_3^{2-}]=K _{sp} $ of $ BaCO_3 $
$ [B{a^{2+}}]=\frac{K _{sp}}{[CO_3^{2-}]}=\frac{5.1\times {10^{-9}}}{1.0\times {10^{-4}}} $
$ =5.1\times {10^{-5}}M $
BETA
