Solid $\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2$ 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 ?

$\left({\mathrm{K}}_{\mathrm{SP}} \mathrm{for} {\mathrm{BaCO}}_3 = 5.1 \times {10}^{-9}\right)$

${\mathrm{Na}}_2{\mathrm{CO}}_3 \rightleftharpoons 2{\mathrm{Na}}^{+} + {\mathrm{CO}}_3^{2-}$

$\left[{\mathrm{Na}}_2{\mathrm{CO}}_3\right] = \left[{\mathrm{CO}}_3^{2-}\right] = 1 \times {10}^{-4} \mathrm{M}$

${\mathrm{BaCO}}_3 \rightleftharpoons {\mathrm{Ba}}^{2+} + {\mathrm{CO}}_3^{2-}$

${\mathrm{K}}_{\mathrm{SP}} \left({\mathrm{BaCO}}_3\right) = \left[{\mathrm{Ba}}^{2+}\right] \left[{\mathrm{CO}}_3^{2-}\right]$

$\left[{\mathrm{Ba}}^{2+}\right] = \frac{{\mathrm{K}}_{\mathrm{SP}} \left({\mathrm{BaCO}}_3\right)}{\left[{\mathrm{CO}}_3^{2-}\right]}$

$\left[{\mathrm{Ba}}^{2+}\right] = \frac{5.1 \times {10}^{-9}}{1 \times {10}^{-4}}$

$\left[{\mathrm{Ba}}^{2+}\right] = 5.1 \times {10}^{-5} \mathrm{M}$

Hence, at  $5.1 \times {10}^{-5}$ concentration of  ${\mathrm{Ba}}^{2+},$ a precipitate will begin to form.