Question

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\operatorname{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Mg} \mathrm{I}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\operatorname{AlBr}_{3}(a q)\)

Short answer

a. A precipitate of Cu(Hg2)(SO4) will form b. No precipitate will form c. A precipitate of MgCO3 will form d. No precipitate will form

Step-by-step solution

Identify ions present

In the first solution, we have Hg2 + ions and NO3 - ions. In the second solution, we have Cu2 + ions and SO4 2- ions.

Formulate possible double replacement reactions

Possible combinations are Hg2 + CuSO4 -> Cu(Hg2)(SO4), and NO3 - + CuSO4 -> Cu(NO3)2.

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that Cu(NO3)2 is soluble in water and no precipitate will form. However, Cu(Hg2)(SO4) is insoluble in water, meaning that a precipitate will form. b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\)

Identify ions present

In the first solution, we have Ni2+ ions and NO3- ions. In the second solution, we have Ca2+ ions and Cl- ions.

Formulate possible double replacement reactions

Possible combinations are Ni(NO3)2 + CaCl2 -> Ca(NO3)2 + NiCl2.

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that both Ca(NO3)2 and NiCl2 are soluble in water, meaning no precipitate will form. c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Mg} \mathrm{I}_{2}(a q)\)

Identify ions present

In the first solution, we have K+ ions and CO3 2- ions. In the second solution, we have Mg2+ ions and I- ions.

Formulate possible double replacement reactions

Possible combinations are K2CO3 + MgI2 -> MgCO3 + 2KI.

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that MgCO3 is insoluble in water, meaning a precipitate will form. KI is soluble in water, so no precipitate will form for that compound. d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\operatorname{AlBr}_{3}(a q)\)

Identify ions present

In the first solution, we have Na+ ions and CrO4 2- ions. In the second solution, we have Al3+ ions and Br- ions.

Formulate possible double replacement reactions

Possible combinations are Na2CrO4 + AlBr3 -> Al(CrO4)3 + 3NaBr.

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that NaBr and Al(CrO4)3 are soluble in water, and no precipitate will form. #Summary# a. A precipitate of Cu(Hg2)(SO4) will form b. No precipitate will form c. A precipitate of MgCO3 will form d. No precipitate will form

Key concepts

Solubility Rules

Understanding solubility rules is crucial when predicting whether a precipitation reaction will occur. Solubility rules are guidelines that chemists use to predict the solubility of ionic compounds in water. These rules help determine if a solid will form when solutions are mixed. In general, most nitrates (\(\text{NO}_3^-\)) and salts containing alkali metal ions, such as sodium (\(\text{Na}^+\)) or potassium (\(\text{K}^+\)), are soluble in water, meaning they dissolve completely and do not precipitate.
On the other hand, most carbonates (\(\text{CO}_3^{2-}\)) and phosphates (\(\text{PO}_4^{3-}\)) are insoluble, unless paired with alkali metals or certain other ions. Sulfates like \(\text{SO}_4^{2-}\) are generally soluble, but they form exceptions with ions like \(\text{Pb}^{2+}\), \(\text{Ba}^{2+}\), and \(\text{Hg}_2^{2+}\), where they become insoluble and precipitate.
To apply these rules, one must list the ions present in the solution and cross-reference them with solubility rules to see if any combination forms an insoluble compound—with such compounds causing precipitate formation.

Double Replacement Reactions

Double replacement reactions, also known as double displacement reactions, involve the exchange of ions between two compounds in solution. The general form for these reactions is:

• \( \text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB} \)

Here, \(\text{A}\) and \(\text{C}\) are cations, while \(\text{B}\) and \(\text{D}\) are anions.
The key to predicting whether a precipitation (or insoluble product) will form in a double replacement reaction is to identify the cations and anions in each compound, then propose new compounds by exchanging partners. These new compounds are then evaluated against solubility rules to determine if they are soluble in water or not.
For instance, mixing \(\text{K}_2\text{CO}_3\) and \(\text{MgI}_2\) would form \(\text{MgCO}_3\) and \(\text{KI}\). According to solubility rules, \(\text{MgCO}_3\) is insoluble, so it forms a precipitate.

Ion Identification

Ion identification is the first step in predicting the outcome of a chemical reaction, especially precipitation reactions. It involves identifying the individual ions that make up the original chemical compounds. In aqueous solutions, ionic compounds dissociate into their respective cations and anions.
For example, the compound \(\text{Na}_2\text{CrO}_4\) dissociates into two \(\text{Na}^+\) ions and one \(\text{CrO}_4^{2-}\) ion in solution. Similarly, \(\text{AlBr}_3\) dissociates into \(\text{Al}^{3+}\) and three \(\text{Br}^-\) ions.
Understanding what ions are present allows chemists to predict how these ions will interact with others in the mixture. It is essential for determining possible product pairs in double replacement reactions and for applying solubility rules effectively.

Chemical Equation Balancing

Balancing chemical equations is crucial to accurately representing the specific chemical reaction taking place, ensuring that the law of conservation of mass is upheld. This means the number of each type of atom must be the same on both sides of a chemical equation.
To balance an equation, one must adjust the coefficients (whole numbers placed before compounds) to reflect the number of molecules or formula units. For example, consider the reaction between \(\text{Na}_2\text{CrO}_4\) and \(\text{AlBr}_3\). The balanced reaction, \(\text{Al(CrO}_4)_3 + 3\text{NaBr}\), shows equal numbers of each kind of atom on both the reactant and product sides.
Balancing equations ensures the reaction adheres to the chemical standards of stoichiometry, allowing for accurate predictions of reaction outcomes, including the formation of precipitates in double replacement reactions.