Solid $Ba {({NO}_{3})}_{2}$ )i is gradually dissolved in a $1.0 \times 10^{- 4} M {Na}_{2} {CO}_{3}$ . solution. At what concentration of ${Ba}^{2 +}$ will a precipitate begin to form? (Given: $K_{sp} = 5.1 \times 10^{- 9} M^{2} for {BaCO}_{3} .$ .)

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

$8.1 \times 10^{- 8} M$

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

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

answer is D.

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Detailed Solution

We have ${BaCO}_{3} (s) ⇌ {Ba}^{2 +} (aq) + {CO}_{3}^{2 -} (aq); K_{sp} = [{Ba}^{2 +}] [{CO}_{3}^{2 -}]$

The minimum [Ba ²⁺] required to start precipitation of BaCO₃ is

$[{Ba}^{2 +}] = \frac{K_{sp}}{[{CO}_{3}^{2 -}]} = \frac{5.1 \times 10^{- 9} M^{2}}{1.0 \times 10^{- 4} M} = 5.1 \times 10^{- 5} M$

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