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Q.

Solid $B a {(N O_{3})}_{2}$ is gradually dissolved in a $1 \times 10^{- 4} M, N a_{2} C O_{3}$ solution. At which concentration of $B a^{+ 2},$ precipitate of $B a C O_{3}$ begins to form ? (Ksp for $B a C O_{3}$ is $5.1 \times 10^{- 9}$ )

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a

$7.1 \times 10^{- 8} M$

b

$5.1 \times 10^{- 5} M$

c

$8.1 \times 10^{- 7} M$

d

$4.1 \times 10^{- 5} M$

answer is A.

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Detailed Solution

$[C {O_{3}}^{2 -}] = 1 \times 10^{- 4} M$
$\Rightarrow S = 1 \times 10^{- 4} M$
At equilibrium, $[B a^{+ 2}] [C {O_{3}}^{2 -}] = K_{S_{p}}$ of $B a C o_{3}$
$∴ [B a^{+ 2}] = \frac{K_{S_{p}}}{[C {o_{3}}^{2 -}]} = \frac{5.1 \times 10^{- 9}}{10^{- 4}} = 5.1 \times 10^{- 5} M$

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Solid Ba(NO3)2 is gradually dissolved in a 1×10−4M,Na2CO3 solution. At which concentration of Ba+2, precipitate of BaCO3 begins to form ? (Ksp forBaCO3 is 5.1×10−9 )