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Solubility Product

The amount of...

Question

The amount of $B a S O_{4}$ formed upon mixing $100$ mL of $20.8 %$ $B a C l_{2}$ solution with $50$ mL of $9.8 %$ $H_{2} S O_{4}$ solution will be:
[Ba $= 137$ , Cl $= 35.5$ , S $= 32$ , H $= 1$ and O $= 16$ ].

23.3

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11.8

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30.6

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33.2

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Solution

The correct option is A $11.8$ g
$H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l$

$\to W / W 20.8$ solution $\Rightarrow \frac{20.8 g B a S O_{4}}{89.2 g w a t e r}$

In $89.2 g$ water, $20.8 g B a S O_{4}$

In $100 g (100 m L)$ water, $x g B a S O_{4}$

$x = \frac{100 \times 20.8}{89.2} = 23.32 g B a S O_{4}$

Number of moles of $B a S O_{4} = \frac{23.32}{M o l a r m a s s o f B a S O_{4}} = \frac{23.32}{233} = 0.1 m o l e s$

$\to W / W 9.8$ solution $\Rightarrow \frac{9.8 g H_{2} S O_{4}}{90.2 g H_{2} O}$

In $90.2 g$ water, $9.8 g H_{2} S O_{4}$

In $50 g (50 m L)$ water, $y g H_{2} S O_{4}$

$y = \frac{50 \times 9.8}{90.2} = 5.43 g H_{2} S O_{4}$

Number of moles of $H_{2} S O_{4} = \frac{5.43}{m o l a r m a s s o f H_{2} S O_{4}} = \frac{5.43}{98} = 0.055 m o l e s$

Since $H_{2} S O_{4}$ is in lesser quantity, it will act as a limiting reagent.

$0.055$ moles of $H_{2} S O_{4}$ will react with $0.055$ moles of $B a C l_{2}$ to give $0.055$ moles of $B a S O_{4}$ according to balanced reaction equation.

$0.055$ moles of $B a S O_{4} = z g r a m s B a S O_{4}$

$1 m o l e B a S O_{4} = 233 g$

$z = \frac{233 \times 0.055}{1} = 11.8 g$

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