Solid Ba(NO3)2 is gradually dissolved in a 23 4 NaCO10M.01 -× solution. At what concentration of Ba2+ will a precipitate begin to form ? )

To determine the concentration of Ba²⁺ at which a precipitate begins to form when solid Ba(NO₃)₂ is dissolved in a sodium carbonate (Na₂CO₃) solution, we need to consider the solubility product constant (K_sp) for barium carbonate (BaCO₃). This is the compound that will precipitate out of solution when the concentrations of Ba²⁺ and CO₃^2- exceed a certain threshold.

Understanding the Reaction

When Ba(NO₃)₂ dissolves in water, it dissociates into Ba²⁺ ions and nitrate ions (NO₃^-). The relevant reaction for precipitation is:

• Ba²⁺ (aq) + CO₃^2- (aq) ⇌ BaCO₃ (s)

Finding the Ksp Value

The K_sp for BaCO₃ at room temperature is approximately 5.0 x 10^-9. This value indicates the product of the concentrations of the ions in a saturated solution:

• K_sp = [Ba²⁺][CO₃^2-]

Calculating the Concentration of Ba²⁺

To find the concentration of Ba²⁺ at which precipitation begins, we need to know the concentration of CO₃^2- in the solution. Assuming you have a 0.01 M Na₂CO₃ solution, the concentration of carbonate ions will also be 0.01 M.

Now, we can set up the equation using the K_sp value:

• 5.0 x 10^-9 = [Ba²⁺][0.01]

Solving for Ba²⁺ Concentration

Rearranging the equation to solve for [Ba²⁺], we get:

• [Ba²⁺] = K_sp / [CO₃^2-]
• [Ba²⁺] = (5.0 x 10^-9) / (0.01)
• [Ba²⁺] = 5.0 x 10^-7 M

Conclusion

Therefore, a precipitate of BaCO₃ will begin to form when the concentration of Ba²⁺ reaches approximately 5.0 x 10^-7 M in the solution. This calculation highlights the importance of understanding solubility products and how they govern the formation of precipitates in chemical reactions.