Question

What percentage of the \(\mathrm{Ba}^{2+}\) in solution is precipitated as \(\mathrm{BaCO}_{3}(\mathrm{s})\) if equal volumes of \(0.0020 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) and \(0.0010 \mathrm{M} \mathrm{BaCl}_{2}(\text { aq })\) are mixed?

Short answer

100% of the \(Ba^{2+}\) ions precipitate as \(BaCO_{3}\) when equal volumes of 0.0020 M \(Na_2CO_3\) and 0.0010 M \(BaCl_2\) are mixed.

Step-by-step solution

Determination of the limiting reactant

First, calculate the amount of moles for each reactant. The number of moles are given by the product of volume and molarity. For \(Ba^{2+}\), it is \(0.0010 M * V\), and for \(CO_3^{2-}\) , it is \(0.0020 M * V\). Since the volumes are equal and cancel out, it is clear that there are fewer moles of \(Ba^{2+}\) present, making it the limiting reactant.

Calculation of the amount of product formed

Next determine the amount of \(BaCO_3\) precipitate formed by using the stoichiometric ratio from the balanced equation. One mole of \(Ba^{2+}\) yields one mole of \(BaCO_3\). Therefore, the amount of \(BaCO_3\) formed is equal to the amount of \(Ba^{2+}\) ions provided, which is \(0.0010 M * V\).

Calculation of the percentage of precipitated \(Ba^{2+}\) ions

Lastly, to find out the percentage of \(Ba^{2+}\) that precipitate as \(BaCO_3\), divide the number of moles of \(BaCO_3\) formed by the total initial moles of \(Ba^{2+}\), and multiply by 100. The total initial moles of \(Ba^{2+}\) is \(0.0010 M * V\). Therefore, the percentage equals to \(\frac{0.0010 M * V }{0.0010 M * V}\ * 100\%\), which is 100%.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is crucial in predicting the outcome of chemical reactions, especially in precipitation reactions like the textbook exercise example. A limiting reactant is the substance in a chemical reaction that is completely consumed when the chemical reaction is complete.

The amount of product formed in a reaction is determined by this limiting reactant because no additional product can form once it has been used up. In our example, we took into consideration the molarity and volume of the reactants to identify that \(Ba^{2+}\) ions were the limiting reactant. This conclusion was reached because, despite equal volumes of solution, there were fewer moles of \(Ba^{2+}\) available compared to \(CO_3^{2-}\) ions.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's based on the balanced chemical equation and allows us to calculate the masses, volumes, and moles of compounds involved in a reaction.

In the context of our problem, stoichiometry helps us determine the exact amount of \(BaCO_3\) that will form, given the amount of limiting reactant available. By using the stoichiometric coefficients from the balanced equation, which tells us that one mole of \(Ba^{2+}\) will react with one mole of \(CO_3^{2-}\) to form one mole of \(BaCO_3\), we are able to predict the formation of \(BaCO_3\) precisely.

Molarity

Molarity, measured in moles per liter (M), is a way to express the concentration of a solution. In our exercise, molarity is used to communicate how much of a substance is present in a given volume of liquid.

When the molarities of two solutions that will react together are known, as in the example of \(0.0020 M \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) and \(0.0010 M \mathrm{BaCl}_{2}(\text{aq})\), it allows us to calculate the number of moles reacting and thereby determine the limiting reactant. Molarity is a foundation for stoichiometric calculations and is integral to solving many chemistry problems, including the percentage of a substance precipitated in a reaction.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products. It's essential to realize, however, that chemical equilibrium is not a factor in the textbook exercise since the formation of a precipitate represents the completion of the reaction.

Nevertheless, understanding chemical equilibrium is important when dealing with reactions where reactants and products are in a dynamic exchange. While it does not directly apply to the determination of the percentage of \(Ba^{2+}\) precipitated in the given problem, chemical equilibrium concepts are vital in more complex scenarios where reversible reactions and their equilibrium constants are taken into account.