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Stoichiometry

The solubilit...

Question

The solubility product of $B a S O_{4}$ at $25^{\circ} C$ is $1.0 \times 10^{- 9}$ . What would be

the concentration of $H_{2} S O_{4}$ necessary to precipitate $B a S O_{4}$ from a solution

of $0.01 M$ $B a^{2 +}$ ions ?

10^-9

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10^-8

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10^-7

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10^-6

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Solution

The correct option is C
10^-7

$B a S O_{4} ⇋ B a^{2 +} + S O_{4}^{2 -}$

At equilibrium, $[B a^{2 +}]$ = $0.01$ , $[S O_{4}^{2 -}]$ = $s$

$K_{s p} = [B a^{2 +}] [S O_{4}^{2 -}] = 0.01 \times s$

$S (S O_{4}^{2 -}) = \frac{K_{s} p}{[B a^{2 +}]} = \frac{1 \times 10^{- 9}}{0.01} = 10^{- 7} m o l e / l i t r e$

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Similar questions

Q. The solubility product of

B a S O_{4}

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1.0 \times 10^{- 9}

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The solubility product of $B a S O_{4}$ is $1.5 \times 10^{- 9}$ . The precipitation in a $0.01 M$ $B a^{2 +}$ ions solution will start on adding $H_{2} S O_{4}$ of concentration -

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The solubility product of BaSO4 at 25∘ C is 1.0 × 10−9. What would be the concentration of H2SO4 necessary to precipitate BaSO4 from a solution of 0.01M Ba2+ ions ?

The correct option is C 10-7 BaSO4 ⇋ Ba2+ + SO2−4 At equilibrium, [Ba2+] = 0.01, [SO2−4] = s Ksp = [Ba2+][SO2−4] = 0.01 × s S(SO2−4) = Ksp[Ba2+] = 1 × 10−90.01 = 10−7 mole/litre

The solubility product of $B a S O_{4}$ at $25^{\circ} C$ is $1.0 \times 10^{- 9}$ . What would be

the concentration of $H_{2} S O_{4}$ necessary to precipitate $B a S O_{4}$ from a solution

of $0.01 M$ $B a^{2 +}$ ions ?