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Chapter 16: Problem 61

If \(20.0 \mathrm{~mL}\) of \(0.10 \mathrm{MBa}\left(\mathrm{NO}_{3}\right)_{2}\) are added to \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) will \(\mathrm{BaCO}_{3}\) precipitate?

Short Answer

Expert verified
To determine whether BaCO3 will precipitate or not, one must compare Q (ion product) and Ksp (solubility product constant) for BaCO3.

Step by step solution

01

Write the Reaction

Write the balanced chemical equation for the reaction. For the reaction of Barium Nitrate (Ba(NO3)2) and Sodium Carbonate (Na2CO3), the products are Sodium Nitrate (NaNO3) and Barium Carbonate (BaCO3). The sensitive to precipitation process is the reaction of Barium Ions (Ba2+) and Carbonate ions (CO32-) producing Barium Carbonate (BaCO3): \[Ba^{2+} + CO3^{2-} \rightarrow BaCO3\]
02

Initial Concentrations

Calculate the initial concentrations of Ba2+ and CO32-. The initial molarity is given, but since the volumes of the solutions are different, the initial number of moles needs to be calculated. For this, multiply the volume of each solution (converted to liters) by its molarity. That will give the amount of moles. Then, divide this by the total volume to find the new concentration.
03

Calculate Ion Product

Calculate the ion product (Q) of Ba2+ and CO32-. The ion product, Q, is found by multiplying the concentrations of the products.
04

Compare Q and Ksp

Compare the ion product Q with the solubility product constant, Ksp, for BaCO3. If Q > Ksp, a precipitate will form, which in this case is our BaCO3. If Q = Ksp, the solution is saturated, and if Q < Ksp , the solution is unsaturated, and there's no precipitation occured.

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