Problem 47
Question
Write equations for the salt hydrolysis reactions occuring when the following salts dissolve in water. Classify each as acidic, basic, or neutral. \begin{equation} \begin{array}{l}{\text { a. ammonium nitrate } \quad \text { c. rubidium acetate }} \\ {\text { b. potassium sulfate}} \quad\quad {\text {d. calcium carbonate }}\end{array} \end{equation}
Step-by-Step Solution
Verified Answer
a. Acidic; b. Neutral; c. Basic; d. Basic.
1Step 1: Understand Salt Hydrolysis
Salt hydrolysis involves the reaction of a salt with water to form an acidic or basic solution. It's important to determine if the resulting solutions are acidic, basic, or neutral by considering the strengths of the parent acids and bases from which the salts are derived.
2Step 2: Determine Hydrolysis and Solution Type for Ammonium Nitrate
Ammonium nitrate (NH₄NO₃) dissolves in water to form ammonium ions (NH₄⁺) and nitrate ions (NO₃⁻). The ammonium ion is the conjugate acid of a weak base (NH₃), leading to the following hydrolysis reaction: \[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \] This produces hydronium ions, making the solution acidic.
3Step 3: Determine Hydrolysis and Solution Type for Potassium Sulfate
Potassium sulfate (K₂SO₄) dissociates completely into potassium ions (K⁺) and sulfate ions (SO₄²⁻). Potassium ion is derived from a strong base (KOH) and sulfate ion from a strong acid (H₂SO₄), so neither ion hydrolyzes significantly. Thus, the solution remains neutral.
4Step 4: Determine Hydrolysis and Solution Type for Rubidium Acetate
Rubidium acetate (RbCH₃COO) dissociates into rubidium ions (Rb⁺) and acetate ions (CH₃COO⁻). The acetate ion, being the conjugate base of the weak acid acetic acid (CH₃COOH), reacts with water: \[ CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^- \] This increases hydroxide ions, making the solution basic.
5Step 5: Determine Hydrolysis and Solution Type for Calcium Carbonate
Calcium carbonate (CaCO₃) dissociates into calcium ions (Ca²⁺) and carbonate ions (CO₃²⁻). The carbonate ion is the conjugate base of the weak acid carbonic acid (H₂CO₃), leading to hydrolysis: \[ CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \] As this increases hydroxide ions, the solution is basic.
Key Concepts
Acidic SolutionsBasic SolutionsNeutral SolutionsConjugate Acid-Base Pairs
Acidic Solutions
Acidic solutions result when a salt dissolves in water and yields hydronium ions (\(H_3O^+\)) as a product of hydrolysis. This typically happens when the salt contains a cation that is the conjugate acid of a weak base.
For example, ammonium nitrate (\(NH_4NO_3\)) dissolves in water to produce ammonium ions (\(NH_4^+\)). This ammonium ion reacts with water:
For example, ammonium nitrate (\(NH_4NO_3\)) dissolves in water to produce ammonium ions (\(NH_4^+\)). This ammonium ion reacts with water:
- The ammonium ion (\(NH_4^+\)) can donate a proton to water, forming ammonia (\(NH_3\)) and hydronium ion (\(H_3O^+\)).
- The increase in hydronium ions lowers the pH of the solution.
- An acidic solution has a pH less than 7.
Basic Solutions
Basic solutions are formed when a salt yields hydroxide ions (\(OH^-\)) upon dissolving in water. These salts often consist of anions that are conjugate bases of weak acids.
Rubidium acetate (\(RbCH_3COO\)) is an example of a salt resulting in a basic solution. When dissolved, rubidium acetate dissociates into acetate ions (\(CH_3COO^-\)) which react:
Rubidium acetate (\(RbCH_3COO\)) is an example of a salt resulting in a basic solution. When dissolved, rubidium acetate dissociates into acetate ions (\(CH_3COO^-\)) which react:
- Acetate ions accept protons from water, forming acetic acid (\(CH_3COOH\)) and releasing hydroxide ions (\(OH^-\)).
- Basic solutions have a pH greater than 7.
- The solution becomes less acidic and more alkaline as hydroxide ions increase.
- The carbonate ion (\(CO_3^{2-}\)) interacts with water, producing bicarbonate (\(HCO_3^-\)) and hydroxide ions.
Neutral Solutions
Neutral solutions occur when a salt dissolves in water without changing the concentration of hydronium (\(H_3O^+\)) or hydroxide (\(OH^-\)) ions. This is common for salts derived from strong acids and strong bases.
Potassium sulfate (\(K_2SO_4\)) is one such salt that results in a neutral solution. Here's why:
Potassium sulfate (\(K_2SO_4\)) is one such salt that results in a neutral solution. Here's why:
- Neither of its ions—potassium (\(K^+\)) from a strong base (KOH) and sulfate (\(SO_4^{2-}\)) from a strong acid (H₂SO₄)—undergo significant hydrolysis.
- Neither ion substantially alters the pH when dissolved.
- Neutral aqueous solutions typically have a pH of 7.
Conjugate Acid-Base Pairs
The concept of conjugate acid-base pairs is essential in understanding salt hydrolysis. When a salt dissolves in water, it may yield ions that are conjugate acids or bases, playing crucial roles in determining the solution's pH.
A conjugate acid is formed when a base gains a proton, and a conjugate base forms when an acid loses one:
A conjugate acid is formed when a base gains a proton, and a conjugate base forms when an acid loses one:
- The strength of a conjugate acid or base affects whether a salt solution is acidic or basic.
- For instance, the ammonium ion (\(NH_4^+\)) is a conjugate acid of ammonia (\(NH_3\)), a weak base.
- It's a conjugate base of acetic acid (\(CH_3COOH\)), a weak acid, thereby affecting the solution's basicity.
- Predicting the direction of reactions in aqueous solutions.
- Determining whether a solution will be acidic, basic, or neutral based on its components.
Other exercises in this chapter
Problem 45
What is the concentration of a household ammonia cleaning solution if 49.90 \(\mathrm{mL}\) of 0.5900 \(M\) \(\mathrm{HCl}\) is required to neutralize 25.00 \(\
View solution Problem 46
Challenge How many milliliters of 0.500 \(\mathrm{M}\) NaOH would neutralize 25.00 \(\mathrm{mL}\) of 0.100 \(M\) \(\mathrm{H}_{3} \mathrm{PO}_{4} ?\)
View solution Problem 48
Challenge Write the equation for the reaction that occurs in a titration of ammonium hydroxide \(\left(\mathrm{NH}_{4} \mathrm{OH}\right]\) with hydrogen bromid
View solution Problem 50
Explain the difference between the equivalence point and the end point of a titration.
View solution