First slide

Solubility product(KSP)

Question

Solid $Ba {({NO}_{3})}_{2}$ 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 ?
$(K_{SP} for {BaCO}_{3} = 5.1 \times 10^{- 9})$

Easy

A

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

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

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

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

Solution

${Na}_{2} {CO}_{3} ⇌ 2 {Na}^{+} + {CO}_{3}^{2 -}$
$[{Na}_{2} {CO}_{3}] = [{CO}_{3}^{2 -}] = 1 \times 10^{- 4} M$
${BaCO}_{3} ⇌ {Ba}^{2 +} + {CO}_{3}^{2 -}$
$K_{SP} ({BaCO}_{3}) = [{Ba}^{2 +}] [{CO}_{3}^{2 -}]$
$[{Ba}^{2 +}] = \frac{K_{SP} ({BaCO}_{3})}{[{CO}_{3}^{2 -}]}$
$[{Ba}^{2 +}] = \frac{5.1 \times 10^{- 9}}{1 \times 10^{- 4}}$
$[{Ba}^{2 +}] = 5.1 \times 10^{- 5} M$
Hence, at $5.1 \times 10^{- 5}$ concentration of ${Ba}^{2 +},$ a precipitate will begin to form.

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