Solid Ba (NO_3 ), is gradually dissolved in a 1.0 10^ -4 M Na_2 C… - Vidyadip

Question

Solid $\mathrm{Ba}\left(\mathrm{NO}_3\right)$, is gradually dissolved in a $1.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3$ solution. At what concentration of $\mathrm{Ba}^{2+}$ will a precipitate begin to form. $\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\left.\mathrm{BaCO}_3=5.1 \times 10^{-9}\right)$

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Answer

$\text{BaCO}_3(\text{s})\rightleftharpoons\text{Ba}^{2+}+\text{CO}^{2-}_{3}$
$\text{Na}_2\text{CO}_3\xrightarrow{\ \ \ \ \ \ \ \ }2\text{Na}^++\text{CO}^{2-}_3$
$[\text{Na}_2\text{CO}_3]=1.0\times10^{-4}\text{M}$
$\text{K}_{\text{sp}}=[\text{Ba}^{2+}][\text{CO}_3^{2-}]$
$5.1\times10^{-9}=[\text{Ba}^{2+}][1\times10^{-4}\text{M}]$
$\text{[Ba}^{2+}]=\frac{5.1\times10^{-9}}{1\times10^{-4}}$
$=5.1\times10^{-5}\text{M}$
At $5.1 \times 10^{-5} \mathrm{M}$ concentration of $\left[\mathrm{Ba}^{2+}\right], \mathrm{I} . \mathrm{P}\left(\right.$ ionic product) will become equal to $\mathrm{K}_{\mathrm{sp}}$ and precipitate will begin to from.

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