Solid $Ba\left( NO_{3} ight)_{2}$ is gradually dissolved in a $1.0 imes 10^{- 4}$ M $Na_{2}CO_{3}$ solution. At what concentration of $Ba^{2 +}$ will a precipitate begin to form? ($K_{sp}\ for\ {BaCO}_{3} = 5.1 imes 10^{- 9})$

Question

Solid $Ba\left( NO_{3} ight)_{2}$ is gradually dissolved in a $1.0 	imes 10^{- 4}$ M $Na_{2}CO_{3}$ solution. At what concentration of $Ba^{2 +}$ will a precipitate begin to form? ($K_{sp}\ for\ {BaCO}_{3} = 5.1 	imes 10^{- 9})$

Dr. Abubacker · Accepted Answer

(b)
$K_{sp}\left\lbrack BaCO_{3} ightbrack = \left\lbrack Ba^{2 +} ightbrack\lbrack CO_{3}^{2 -}brack$
$\left\lbrack Ba^{2 +} ightbrack = \frac{K_{sp}}{\lbrack CO_{3}^{2 -}brack} = \frac{5.1 	imes 10^{- 9}}{1 	imes 10^{- 4}}$
$\left\lbrack Ba^{2 +} ightbrack = 5.1 	imes 10^{- 5}\ M$

Q. Solid $Ba\left( NO_{3} \right)_{2}$ 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? ($K_{sp}\ for\ {BaCO}_{3} = 5.1 \times 10^{- 9})$

Explanation

Reviewed by Dr. Abubacker

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