A) \[\frac{{{y}^{2}}}{{{({{y}_{1}}+{{y}_{2}}-{{y}_{3}})}^{2}}}\]
B) \[\frac{2.5{{y}^{-2}}}{{{({{y}_{1}}+{{y}_{2}}-{{y}_{3}})}^{2}}}\]
C) \[\frac{500}{{{y}_{1}}+{{y}_{2}}-{{y}_{3}}}\]
D) \[\frac{2.5\times {{10}^{5}}{{y}^{2}}}{{{({{y}_{1}}+{{y}_{2}}-{{y}_{3}})}^{2}}}\]
Correct Answer: D
Solution :
| [d] \[\Delta _{m}^{\infty }BaS{{O}_{4}}=2\Delta _{eq}^{\infty }(BaS{{O}_{4}})\] |
| \[\Delta _{eq}^{\infty }(BaS{{O}_{4}})=\Delta _{eq}^{\infty }(B{{a}^{2+}})+\Delta _{eq}^{\infty }(SO_{4}^{-2})\] |
| \[=\Delta _{eq}^{\infty }(BaC{{l}_{2}})+\Delta _{eq}^{\infty }({{H}_{2}}S{{O}_{4}})-\Delta _{eq}^{\infty }(HCl)\] |
| \[\Delta _{eq}^{\infty }(BaS{{O}_{4}}){{y}_{1}}+{{y}_{2}}-{{y}_{3}}\] |
| \[\Delta _{m}^{\infty }=2({{y}_{1}}+{{y}_{2}}-{{y}_{3}})\] |
| For sparingly soluble salt, |
| \[\Delta _{m}^{\infty }=\frac{k}{m}\times 100\] |
| or \[M=\frac{y}{2({{y}_{1}}+{{y}_{2}}-{{y}_{3}})}\times 1000\] |
| \[=\frac{500}{{{y}_{1}}+{{y}_{2}}-{{y}_{3}}}\] |
| \[{{K}_{sp}}={{M}^{2}}=\frac{2.5\times {{10}^{5}}{{y}^{2}}}{{{({{y}_{1}}+{{y}_{2}}-{{y}_{3}})}^{2}}}\] |
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