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:

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

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

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