| The solubility of \[\text{BaS}{{\text{O}}_{\text{4}}}\] in water is \[\text{2}\text{.42 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--3}}}\text{g}{{\text{L}}^{\text{-1}}}\] at 298 K. The value of its solubility product\[\text{(}{{\text{K}}_{\text{sp}}}\text{)}\] will be [NEET - 2018] |
| (Given molar mass of\[\text{BaS}{{\text{O}}_{\text{4}}}\text{=233 g mo}{{\text{l}}^{\text{--1}}}\]) |
A) \[\text{1}\text{.08 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--14}}}\text{ mol 2}{{\text{L}}^{\text{--2}}}\]
B) \[\text{1}\text{.08 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--12}}}\text{mo}{{\text{l}}^{\text{2}}}{{\text{L}}^{\text{--2}}}\]
C) \[\text{1}\text{.08 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--10}}}\text{mo}{{\text{l}}^{\text{2}}}{{\text{L}}^{\text{--2}}}\]
D) \[\text{1}\text{.08 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--8}}}\text{mo}{{\text{l}}^{\text{2}}}{{\text{L}}^{\text{--2}}}\]
Correct Answer: C
Solution :
| [c] Solubility of |
| \[\text{BaS}{{\text{O}}_{\text{4}}}\text{,s=}\frac{\text{2}\text{.42 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}}{\text{233}}\text{(mol}\,\,{{\text{L}}^{\text{-1}}}\text{)}\] |
| \[\text{=1}\text{.04 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{(mol}\,\,\text{L-1)}\] |
| \[\text{BaS}{{\text{O}}_{4}}(s)\underset{\text{S}}{\mathop{B{{a}^{2+}}}}\,(aq)+\underset{\text{S}}{\mathop{SO_{4}^{2-}}}\,(aq)\] |
| \[{{\text{K}}_{\text{sp}}}\text{= }\!\![\!\!\text{ B}{{\text{a}}^{\text{2+}}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ SO}_{\text{4}}^{\text{2-}}\text{ }\!\!]\!\!\text{ =}{{\text{s}}^{\text{2}}}\] |
| \[\text{=(1}\text{.04}\times \text{1}{{\text{0}}^{-5}}{{)}^{2}}\] |
| \[=1.08\times {{10}^{-10}}mo{{l}^{2}}{{L}^{-2}}\] |
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