JEE Main On 08 April 2018 Question 17
Question: An aqueous solution contains an unknown concentration of $ B{a^{\text{2+}}} $ . When 50 mL of a 1 M solution of $ N{a_2}S{O_4} $ is added, $ BaS{O_4} $ just begins to precipitate. The final volume is 500 mL. The solubility product of $ BaS{O_4} $ is $ 1\times {10^{-10}} $ . What is the original concentration of $ B{a^{2+}}? $ [JEE Main Online 08-04-2018]
Options:
A) $ 1.1\times {10^{-9}}M $
B) $ 1.0\times {10^{-10}}M $
C) $ 5\times {10^{-9}}M $
D) $ 2\times {10^{-9}}M $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ BaS{O_4}\text{(s)}B{a^{\text{2+}}}\text{(aq)+S}O_4^{\text{2-}}\text{(aq)} $ $ N{a_2}S{O_4}\text{(aq)}\xrightarrow{{}}2N{a^{\text{2+}}}\text{(aq)+S}O_4^{\text{2-}}\text{(aq)} $ $ \frac{1\times 50}{500}=0.1M $ $ {K _{sp}}\text{= }[\text{ B}{a^{\text{2+}}}{] _{final}}{{[\text{ S}O_4^{\text{2-}}]} _{final}} $ $ 1\times {10^{-10}}=( {{[B{a^{+2}}]} _{initial}}\times \frac{450}{500} )\times [0.1] $ $ [\text{ B}{a^{\text{2+}}}]\text{ =1}\text{.1 }\times\text{ 1}{0^{\text{-9}}}M $