NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Ionic Equilibrium
NTA Abhyas 2020 · Chemistry · Ionic Equilibrium
Solid Na _ 2 SO _ 4 is slowly added to a solution which is 0.020M...
Solid is slowly added to a solution which is in _{2}0.020MPb{\left({NO}_{3}\right)}_{2}{Na}_{2}{SO}_{4}{\text{Ba}^{\text{2+}}{\text{PbSO}_{4}{\left[K\right\}_{sp}\left({BaSO}_{4}\right) = 1.0 \times {10}^{- 10}{K}_{sp} \left({PbSO}_{4}\right) =1.6 \times{10}^{- 8}]
Options
- A& 5.0 × 10^{-9} M
- B& 8.0 × 10^{-7} M
- C& 1.25 × 10^{-4}M
- D& & 1.95 × 10^{-8}M
Correct answer
C. & 1.25 × 10^{-4}M
Step-by-step solution
When begins to precipitate
\left[{SO}_{4}^{2 -}\right] = \dfrac{{K}_{sp}\left[BaS{O}_{4}\right]}{\left[Ba^{2 +}\right]} = \dfrac{1.0 \times 1{0}^{- 10}}{0.020}= 5.0 \times {10}^{- 9}M{PbSO}_{4}begins to precipitate<br /><br />\left[{SO}_{4}^{2 -}\right] = \dfrac{{K}_{sp}\left(\text{ PbS }{O}_{4}\right)}{\left[Pb^{2 +}\right]} = \dfrac{1.6 \times 1{0}^{- 8}}{0.020}= 8.0 \times {10}^{- 7}Mis less for precipitation so precipitates first when begins to precipitate \left[{SO}_{4}^{2 -}\right] = 8.0 \times {10}^{- 7}M{Ba}^{2 +} = \dfrac{{K}_{sp}\left(BaS{O}_{4}\right)}{\left[S{O}_{4}^{2 -}\right]}= \dfrac{1.0 \times 1{0}^{- 10}}{8.0 \times 1{0}^{- 7}} = 1.25 \times {10}^{- 4}M
\left[{SO}_{4}^{2 -}\right] = \dfrac{{K}_{sp}\left[BaS{O}_{4}\right]}{\left[Ba^{2 +}\right]} = \dfrac{1.0 \times 1{0}^{- 10}}{0.020}= 5.0 \times {10}^{- 9}M{PbSO}_{4}begins to precipitate<br /><br />\left[{SO}_{4}^{2 -}\right] = \dfrac{{K}_{sp}\left(\text{ PbS }{O}_{4}\right)}{\left[Pb^{2 +}\right]} = \dfrac{1.6 \times 1{0}^{- 8}}{0.020}= 8.0 \times {10}^{- 7}Mis less for precipitation so precipitates first when begins to precipitate \left[{SO}_{4}^{2 -}\right] = 8.0 \times {10}^{- 7}M{Ba}^{2 +} = \dfrac{{K}_{sp}\left(BaS{O}_{4}\right)}{\left[S{O}_{4}^{2 -}\right]}= \dfrac{1.0 \times 1{0}^{- 10}}{8.0 \times 1{0}^{- 7}} = 1.25 \times {10}^{- 4}M