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. a. ammonium nitrate c. rubidium acetate b. potassium sulfate d. calcium carbonate
Step-by-Step Solution
Verified Answer
a. Ammonium nitrate: acidic; hydrolysis reaction: \(NH₄⁺_{(aq)} + H₂O_{(l)} \rightleftharpoons NH₄OH_{(aq)} + H₃O⁺_{(aq)}\)
b. Potassium sulfate: neutral; no hydrolysis reaction
c. Rubidium acetate: basic; hydrolysis reaction: \(CH₃COO⁻_{(aq)} + H₂O_{(l)} \rightleftharpoons CH₃COOH_{(aq)} + OH⁻_{(aq)}\)
d. Calcium carbonate: basic; hydrolysis reaction: \(CO₃²⁻_{(aq)} + H₂O_{(l)} \rightleftharpoons HCO₃⁻_{(aq)} + OH⁻_{(aq)}\)
1Step 1: Identify the parent acid and base for each salt
For each salt, we will first identify the parent acid and base that formed the salt.
a. ammonium nitrate: NH₄⁺ (from NH₄OH, a weak base) and NO₃⁻ (from HNO₃, a strong acid)
b. potassium sulfate: K⁺ (from KOH, a strong base) and SO₄²⁻ (from H₂SO₄, a strong acid)
c. rubidium acetate: Rb⁺ (from RbOH, a strong base) and CH₃COO⁻ (from CH₃COOH, a weak acid)
d. calcium carbonate: Ca²⁺ (from Ca(OH)₂, a strong base) and CO₃²⁻ (from H₂CO₃, a weak acid)
2Step 2: Write the hydrolysis reactions for each salt
Now, we will write the hydrolysis reactions for each salt.
a. Ammonium nitrate:
\(NH₄⁺_{(aq)} + H₂O_{(l)} \rightleftharpoons NH₄OH_{(aq)} + H₃O⁺_{(aq)}\)
NO₃⁻ does not react, as it is a spectator ion.
b. Potassium sulfate:
K⁺ and SO₄²⁻ ions do not react in this case, as they both come from strong acids and bases. The salt does not undergo hydrolysis and do not change the pH of the solution.
c. Rubidium acetate:
\(CH₃COO⁻_{(aq)} + H₂O_{(l)} \rightleftharpoons CH₃COOH_{(aq)} + OH⁻_{(aq)}\)
Rb⁺ does not react, as it is a spectator ion.
d. Calcium carbonate:
\(CO₃²⁻_{(aq)} + H₂O_{(l)} \rightleftharpoons HCO₃⁻_{(aq)} + OH⁻_{(aq)}\)
Ca²⁺ does not react, as it is a spectator ion.
3Step 3: Classify each salt as acidic, basic, or neutral
Based on the hydrolysis reactions, we can classify each salt as acidic, basic, or neutral.
a. Ammonium nitrate: NH₄⁺ produces H₃O⁺ ions in the solution while NO₃⁻ does not react, so the solution is acidic.
b. Potassium sulfate: Neither K⁺ nor SO₄²⁻ ions react in the solution, so the solution is neutral.
c. Rubidium acetate: CH₃COO⁻ produces OH⁻ ions in the solution while Rb⁺ does not react, so the solution is basic.
d. Calcium carbonate: CO₃²⁻ produces OH⁻ ions in the solution while Ca²⁺ does not react, so the solution is basic.
Key Concepts
Acid-Base ChemistryChemical EquationsHydrolysis of SaltspH Classification
Acid-Base Chemistry
Understanding acid-base chemistry is pivotal when analyzing salt hydrolysis reactions. This branch of chemistry involves acids, which donate protons (H⁺), and bases, which accept protons. When salts, which are ionic compounds, dissolve in water, they can undergo hydrolysis, a reaction with water. This reaction depends on the acidic or basic nature of the ions that constitute the salt.
For instance, in our exercise, ammonium nitrate is formed from a weak base (ammonium hydroxide, NH₄OH) and a strong acid (nitric acid, HNO₃). So, when ammonium nitrate is dissolved in water, the ammonium ion (NH₄⁺) can release a proton to form water, resulting in an acidic solution. This is due to the NH₄⁺ coming from a weak base and therefore readily donating its hydrogen ion to water. On the other hand, the nitrate ion (NO₃⁻) does not react and is considered a spectator ion.
For instance, in our exercise, ammonium nitrate is formed from a weak base (ammonium hydroxide, NH₄OH) and a strong acid (nitric acid, HNO₃). So, when ammonium nitrate is dissolved in water, the ammonium ion (NH₄⁺) can release a proton to form water, resulting in an acidic solution. This is due to the NH₄⁺ coming from a weak base and therefore readily donating its hydrogen ion to water. On the other hand, the nitrate ion (NO₃⁻) does not react and is considered a spectator ion.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They show the reactants being transformed into products, with reactant and product formulas balanced to satisfy the law of conservation of mass. In the context of hydrolysis, the chemical equations depict how water interacts with the ions from salts.
For every salt listed in the exercise, we derived equations that illustrate the potential interactions. For example, rubidium acetate, when dissolved in water, is represented by the hydrolysis reaction of its acetate ion with water, forming acetic acid (CH₃COOH) and hydroxide ions (OH⁻). An equation like this helps visualize the hydrolysis process and predict the pH nature of the solution. As the rubidium ion (Rb⁺) is from a strong base and does not react, it's also labeled as a 'spectator ion'.
For every salt listed in the exercise, we derived equations that illustrate the potential interactions. For example, rubidium acetate, when dissolved in water, is represented by the hydrolysis reaction of its acetate ion with water, forming acetic acid (CH₃COOH) and hydroxide ions (OH⁻). An equation like this helps visualize the hydrolysis process and predict the pH nature of the solution. As the rubidium ion (Rb⁺) is from a strong base and does not react, it's also labeled as a 'spectator ion'.
Hydrolysis of Salts
Hydrolysis of salts is a reaction where the anion or cation of the salt reacts with water to produce an acidity or basicity in the solution. When this occurs, salts derived from strong acids and strong bases show no change in the water's pH and are neutral. Conversely, salts from weak acids or bases will react with water, changing the pH.
Acidic Salt Hydrolysis
Ammonium nitrate demonstrates acidic hydrolysis. The ammonium ion (NH₄⁺) reacts with water to form ammonium hydroxide and hydronium ions, contributing to the acidity of the solution.Basic Salt Hydrolysis
On the other hand, calcium carbonate exhibits basic hydrolysis. The carbonate ion (CO₃²⁻) reacts with water to produce bicarbonate (HCO₃⁻) and hydroxide ions, which increase the solution's basicity.pH Classification
pH classification involves categorizing a solution based on its acidity or basicity. This scale ranges from 0 to 14, with values less than 7 indicating acidity, values more than 7 indicating basicity, and a value of 7 indicating neutrality.
In our exercise, we classified each salt solution based on whether the ions produced hydronium (H₃O⁺) or hydroxide (OH⁻) ions during hydrolysis. Ammonium nitrate's hydrolysis leads to an acidic solution because it increases the concentration of hydronium ions. Potassium sulfate remains neutral since neither ion reacts to affect the pH. Rubidium acetate is basic due to the hydroxide ions produced. Similarly, calcium carbonate also results in a basic solution. These classifications help predict the behavior of the compounds in various chemical environments and their potential interactions with other substances.
In our exercise, we classified each salt solution based on whether the ions produced hydronium (H₃O⁺) or hydroxide (OH⁻) ions during hydrolysis. Ammonium nitrate's hydrolysis leads to an acidic solution because it increases the concentration of hydronium ions. Potassium sulfate remains neutral since neither ion reacts to affect the pH. Rubidium acetate is basic due to the hydroxide ions produced. Similarly, calcium carbonate also results in a basic solution. These classifications help predict the behavior of the compounds in various chemical environments and their potential interactions with other substances.
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 \(\mathrm{M} \mathrm{HCl}\) is required to neutralize 25.0
View solution Problem 46
Challenge How many milliliters of 0.500 \(\mathrm{M}\) NaOH would neutralize 25.00 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?\)
View solution Problem 48
Challenge Write the equation for the reaction that occurs in a titration of ammonium hydroxide (NH \(_{4} \mathrm{OH}\) ) with hydrogen bromide \((\mathrm{HBr})
View solution Problem 50
Explain the difference between the equivalence point and the end point of a titration.
View solution