Question:
Solid $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is gradully 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 $\left.\mathrm{Ba} \mathrm{CO}_{3}=5.1 \times 10^{-9}\right)$
Correct Option: , 4
Solution:
$5.1 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right]\left[10^{-4}\right]$
${\left[\mathrm{Ba}^{+2}\right]=5.1 \times 10^{-5} \mathrm{M} }$
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