Solid Ba(NO3)2 is gradually dissolved in a 1 × 10^-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate will form

Dear Student,

We require the K_sp value for BaCO₃ to solve this equation.
The K_sp for BaCO₃ is 5.1 x 10^-9
The reaction can be given as :

Ba(NO₃)₂ + CaCO₃ = BaCO₃ + NaNO₃

Here,
$$\begin{gathered} \left[C{O_3}^{2-}\right]=\left[N{a}_{2}C{O}_3\right]={10}^{-4 }M \\ So, \\ {K}_{sp }=\left[B{a}^{2+}\right]\left[C{O_3}^{2-}\right] = 5.1 x {10}^{-9} \\ \left[B{a}^{2+}\right] x {10}^{-4}= 5.1 x {10}^{-9} \\ So, \\ \left[B{a}^{2+}\right] = 5.1 x {10}^{-5} M \end{gathered}$$

So, precipitation of Ba²⁺ will start at 5.1 x 10^-5 M concentration.

Hope it helps.

Regards