NEET Solved Paper 2018 Question 32
Question: The solubility of $ BaS{O_4} $ in water is $ 2\text{.42 }\times\text{ 1}{0^{\text{–3}}}g{L^{\text{-1}}} $ at 298 K. The value of its solubility product $ \text{(}{K _{sp}}\text{)} $ will be (Given molar mass of $ BaS{O_4}\text{=233 g mo}{l^{\text{–1}}} $ )[NEET - 2018]
Options:
A) $ 1\text{.08 }\times\text{ 1}{0^{\text{–14}}}\text{ mol 2}{L^{\text{–2}}} $
B) $ 1\text{.08 }\times\text{ 1}{0^{\text{–12}}}mo{l^{2}}{L^{\text{–2}}} $
C) $ 1\text{.08 }\times\text{ 1}{0^{\text{–10}}}mo{l^{2}}{L^{\text{–2}}} $
D) $ 1\text{.08 }\times\text{ 1}{0^{\text{–8}}}mo{l^{2}}{L^{\text{–2}}} $
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Answer:
Correct Answer: C
Solution:
- Solubility of $ BaS{O_4}\text{,s=}\frac{2\text{.42 }\times\text{ 1}{0^{\text{-3}}}}{233}\text{(mol}{L^{\text{-1}}}\text{)} $ $ \text{=1}\text{.04 }\times\text{ 1}{0^{\text{-5}}}\text{(mol}\text{L-1)} $ $ BaS{O_4}(s)\underset{S}{\mathop{B{a^{2+}}}},(aq)+\underset{S}{\mathop{SO_4^{2-}}},(aq) $ $ {K _{sp}}\text{= }[\text{ B}{a^{\text{2+}}}]\text{ }[\text{ SO}_4^{\text{2-}}]\text{ =}{s^{2}} $ $ \text{=(1}\text{.04}\times 1{0^{-5}}{{)}^{2}} $ $ =1.08\times {10^{-10}}mol^{2}{L^{-2}} $
