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Common Ion Effect

100 mL of 20....

Question

100 mL of 20.8% $B a C l_{2}$ solution and 50 mL of 9.8% $H_{2} S O_{4}$ solution will form $B a S O_{4}$
$(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)$
$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$

23.3 g

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11.65 g

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30.6 g

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None of these

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Solution

The correct option is D 11.65 g
$100 m l$ of $20.8$ % $B a C l_{2}$ solution= $20.8 g$ $B a C l_{2}$
$50 m l$ of $9.8$ % $H_{2} S O_{4}$ solution= $4.9 g$ $H_{2} S O_{4}$
Reaction: $B a C l_{2} + H_{2} S O_{4} ⟶ B a S O_{4} ↓ + 2 H C l$
$208 g m o l^{- 1}$ $98 g m o l^{- 1}$ $233 g m o l^{- 1}$
$∴ 98 g$ $H_{2} S O_{4}$ reacts with $208 g$ $B a C l_{2}$
$4.9 g$ $H_{2} S O_{4}$ reacts with $\frac{208}{98} \times 4.9 = 10.4 g$ $B a C l_{2}$
$98 g$ $H_{2} S O_{4}$ will produce $233 g$ $B a S O_{4}$
$∴ 4.9 g$ $H_{2} S O_{4}$ will produce= $\frac{233}{98} \times 4.9 = 11.65 g$ $B a S O_{4}$

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100 mL of 20.8% BaCl2 solution and 50 mL of 9.8% H2SO4 solution will form BaSO4(Ba=137,Cl=35.5,S=32,H=1,O=16)BaCl2+H2SO4→BaSO4+2HCl

100 mL of 20.8% $B a C l_{2}$ solution and 50 mL of 9.8% $H_{2} S O_{4}$ solution will form $B a S O_{4}$
$(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)$
$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$