# Solve this following

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 $\left.\mathrm{Ba} \mathrm{CO}_{3}=5.1 \times 10^{-9}\right)$

1. $8.1 \times 10^{-8} \mathrm{M}$

2. $8.1 \times 10^{-7} \mathrm{M}$

3. $4.1 \times 10^{-5} \mathrm{M}$

4. $5.1 \times 10^{-5} \mathrm{M}$

Correct Option: , 4

Solution:

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

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