Question

Write a balanced equation to describe any precipitation reaction that should occur when the following substances are mixed. (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{BaCl}_{2}(a q)\) (b) \(\mathrm{CaS}(a q)\) and \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)(a q)\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{2},_{2}(a q)\right.\) and \(\mathrm{K}_{1} \mathrm{SO}_{4}\) (aq)

Short answer

For reaction (a), the balanced chemical equation is \( \mathrm{K}_{2} \mathrm{CO}_{3}(a q) + \mathrm{BaCl}_{2}(a q) \rightarrow 2 \mathrm{KCl} (aq) + \mathrm{BaCO}_{3}(s) \). For reaction (b), it is \(\mathrm{CaS}(a q) + \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} (aq) \rightarrow \mathrm{HgS}(s) + \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} (aq) \). For reaction (c), it is \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} (aq) + \mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s) + 2\mathrm{KNO}_{3} (aq) \).

Step-by-step solution

Predict the Potential Products

First, you must identify and combine the anion of one reactant with the cation of the other, then do the opposite. For reaction (a), the potential products are KCl and BaCO3. For reaction (b), the potential products are HgS and Ca(NO3)2. For a reaction (c), the products would be PbSO4 and 2 KNO2.

Apply Solubility Rules

Next, check if the potential products are soluble or insoluble using the solubility rules. KCl is soluble, BaCO3 is insoluble, HgS is insoluble, Ca(NO3)2 is soluble, PbSO4 is insoluble and KNO2 is soluble. So, we have predicted the formation of precipitants for each reaction correctly.

Write and Balance the Equations

Write the balanced chemical equations with the correct stoichiometric coefficients and correct states of matter. For reaction (a), the balanced chemical equation is \( \mathrm{K}_{2} \mathrm{CO}_{3}(a q) + \mathrm{BaCl}_{2}(a q) \rightarrow 2 \mathrm{KCl} (aq) + \mathrm{BaCO}_{3}(s) \). For reaction (b), it is \(\mathrm{CaS}(a q) + \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} (aq) \rightarrow \mathrm{HgS}(s) + \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} (aq) \). For reaction (c), it is \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} (aq) + \mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s) + 2\mathrm{KNO}_{3} (aq) \).

Key concepts

Chemical Equation Balancing

Balancing chemical equations is a fundamental aspect of chemistry that ensures the Law of Conservation of Mass is obeyed. This law states that matter cannot be created or destroyed in a chemical reaction. Consequently, the number of each type of atom must be the same on both sides of the equation.

For a balanced equation, count the atoms of each element on both sides and introduce coefficients to even out any discrepancies. In the given exercise, for reaction (a), the balanced equation is \( \mathrm{K}_{2} \mathrm{CO}_{3}(aq) + \mathrm{BaCl}_{2}(aq) \rightarrow 2 \mathrm{KCl} (aq) + \mathrm{BaCO}_{3}(s) \). This equation demonstrates that for every molecule of barium carbonate that precipitates, two molecules of potassium chloride remain in solution.

Solubility Rules

Understanding solubility rules is crucial when predicting the outcome of a chemical reaction, especially when dealing with precipitation reactions. Solubility rules are guidelines that help us predict whether a compound will dissolve (be soluble) or form a precipitate (be insoluble) in water.

Common solubility rules include:

• Most nitrate (\(NO_3^-\)) salts are soluble.
• Sulfates (\(SO_4^{2-}\)) are generally soluble, except those of barium, lead, calcium, and strontium.
• Chlorides (\(Cl^-\)), bromides (\(Br^-\)), and iodides (\(I^-\)) are soluble except when paired with silver, lead, or mercury.

By referring to these rules, we can predict that in reaction (b), \( \mathrm{HgS} \) is the precipitate because sulfides are typically insoluble, except those of the alkali metals and ammonium.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using the coefficients of a balanced chemical equation to calculate the masses, volumes, or moles of substances involved in the reaction.

In the case of reaction (c), stoichiometry tells us that to form lead sulfate \(\mathrm{PbSO}_{4}\) as a precipitate, a 1:1 molar ratio is required when \(\mathrm{Pb(NO}_{3})_{2}\) reacts with \(\mathrm{K}_{2}\mathrm{SO}_{4}\). This means that for every one mole of \(\mathrm{Pb(NO}_{3})_{2}\) used, one mole of \(\mathrm{K}_{2}\mathrm{SO}_{4}\) is required, which results in the formation of one mole of \(\mathrm{PbSO}_{4}\) and two moles of \(\mathrm{KNO}_{3}\) according to the balanced equation. With careful application of stoichiometry, students can predict the amounts of reactants needed and the quantity of product that will be formed.