Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
Precipitate of BaCO₃ will form when 0.96g of Na₂CO₃ is combined with 0.20g BaBr₂ in a 10 L solution.
Precipitate will form in the reaction if for the given reaction value of Qsp i.e. solubility product quotient is more than Ksp i.e. solubility product constant.
Given chemical reaction is:
Na₂CO₃ + BaBr₂ → 2NaBr + BaCO₃
We can calculate the concentration of reactants in terms of molarity, as no. of moles of solute present in per liter of solution.
And moles will be calculate as n = W/M, where
W = given mass
M = molar mass
Given mass of Na₂CO₃ = 0.96g
Moles of Na₂CO₃ = 0.96g / 106 g/mol = 9.1×10⁻³ moles
Volume of solution = 10L
Concentration of Na₂CO₃ = 9.1×10⁻³ moles/10L = 9.1×10⁻⁴ M
Given mass of BaBr₂ = 0.20g
Moles of BaBr₂ = 0.20g / 297g/mol = 6.7×10⁻⁴ moles
Concentration of BaBr₂ = 6.7×10⁻⁴ moles/10L = 6.7×10⁻⁵ M
Qsp will be calculated as the product of their concentration:
Qsp = (9.1×10⁻⁴)(6.7×10⁻⁵) = 6.1×10⁻⁸
And value of Ksp for BaCO₃ = 5.1×10⁻⁹
From this we conclude that BaCO₃ will form precipitate.
Hence, precipitation of BaCO₃ takes place.
To know more about Qsp & Ksp, visit the below link:
https://brainly.com/question/7185695
