Question

A sample of \(20.0 \mathrm{~mL}\) of \(0.10 \mathrm{MBa}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\). Will \(\mathrm{BaCO}_{3}\) precipitate?

Short answer

Yes, BaCO3 will precipitate as the final molarity of BaCO3 in the solution surpasses its solubility limit.

Step-by-step solution

Write the Reaction Equation

The reaction equation would be: \(Ba(NO_3)_2 + Na_2CO_3 \rightarrow 2NaNO_3 + BaCO_3\)

Calculate Moles of Reactants

For \(Ba(NO_3)_2, Number\ of\ moles = Molarity \times Volume = 0.10M \times 20.0mL = 2.0mmol\). For \(Na_2CO_3, Number\ of\ moles = Molarity \times Volume = 0.10M \times 50.0mL = 5.0mmol\)

Check Limiting Reactant and Perform Stoichiometry on Reaction

From the reaction equation, we can see that Ba(NO3)2 and Na2CO3 react in a 1:1 ratio. Since Ba(NO3)2 is the limiting reactant (only 2 mmol present), we can go further and calculate that 2mmol of Na2CO3 will react to form 2mmol of BaCO3

Calculate New Solution Volume and Determine Molarity of BaCO3

The final volume of the solution is the total volume of Ba(NO3)2 and Na2CO3 added together, which is 70mL. The molarity of BaCO3 will be the moles of BaCO3 divided by the new volume, which is \( 2 mmol / 70 mL = 0.0286 M \)

Compare Molarity with Solubility

The solubility of BaCO3 at room temperature is approximately \( 0.0002 M \). Therefore, having a molarity of \(0.0286 M\) exceeds this solubility limit, so BaCO3 will precipitate.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with quantifying the relationships between reactants and products in a chemical reaction. It involves making precise measurements of substances, either before or after a chemical process, to ensure the correct proportions are being used or formed. The basic principle is based on the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

When assessing a reaction, like the dissolution of barium nitrate and sodium carbonate forming sodium nitrate and barium carbonate, stoichiometry is used to calculate the amounts of reactants needed and the quantity of products formed. This is achieved by balancing the reaction equation (Step 1) and then calculating the moles of each reactant (Step 2). In a balanced equation, coefficients indicate the proportion of molecules involved, allowing us to establish molar ratios. These ratios are the heart of stoichiometry and guide us in predicting the outcomes of chemical reactions.

Limiting Reactant

The limiting reactant, or limiting reagent, is the substance that is completely consumed first during a chemical reaction and thus determines the amount of product that can be formed. The opposite, the excess reactant, is the substance that remains after the reaction has reached completion.

Identifying the limiting reactant is a crucial step in many chemistry problems (as highlighted in Step 3). Once the limiting reactant is used up, no further reaction can occur. This concept is akin to a recipe - if you run out of flour while baking cookies, it doesn't matter how much sugar you have left; you can't make more cookies without additional flour. In the given exercise, despite having more moles of sodium carbonate, barium nitrate is the limiting reactant because it has fewer moles available, thereby dictating the maximum amount of barium carbonate that can form.

Solubility Product

The solubility product, symbolized as Ksp, is an equilibrium constant used to explain how saturated solutions of ionic compounds can dissociate into their constituent ions in water. It is a measure of the extent of dissolving that occurs and it applies when the solution is in a dynamic equilibrium, meaning the rate of dissolution is equal to the rate of precipitation.

In simple terms, each substance has a specific solubility product at a given temperature, which can be used to predict whether a precipitate will form. If the ionic product, calculated from the concentrations of the ions in the solution, exceeds Ksp, the solution is oversaturated and precipitation occurs (Step 5). In this exercise, the concentration of barium carbonate exceeds its Ksp, indicating that precipitation is indeed favorable.

Molarity Calculations

Molarity is a unit of concentration expressed as moles of a solute divided by liters of solution. Performing molarity calculations involves quantifying the concentration of reactants or products in a solution, which is essential in predicting the direction and extent of chemical reactions.

In a titration or mixing scenario like the one presented in the problem, after the reaction has occurred, it's important to recount molarity, considering the changes in the volume of the solution (Step 4). These calculations inform us of the concentration of the remaining species and the newly formed compounds. In this example, we calculate the molarity of barium carbonate to compare it with its known solubility and confirm whether a precipitate will form.