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Chapter 13: Problem 117

Determine \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of each of the following solutions. a. \(1.0 M\) KCl b. \(1.0 M \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short Answer

Expert verified
For 1.0 M KCl solution: [OH鈦籡 = \(1 \times 10^{-7}\) M [H鈦篯 = \(1 \times 10^{-7}\) M pH = 7 For 1.0 M KC鈧侶鈧僌鈧� solution: [OH鈦籡 鈮� \(2.36 \times 10^{-12}\) M [H鈦篯 鈮� \(4.24 \times 10^{-3}\) M pH 鈮� 2.37

Step by step solution

01

Determine the nature of the solutions

We need to determine if the given solutions are acidic, basic, or neutral. a. KCl is formed from the reaction between KOH (a strong base) and HCl (a strong acid). Since both are strong, the solution is neutral. b. KC鈧侶鈧僌鈧� is formed from the reaction between KOH (a strong base) and CH鈧僀OOH (a weak acid). Since the base is strong and the acid is weak, the solution is basic.
02

Calculate [OH鈦籡, [H鈦篯, and pH for a neutral solution (1.0 M KCl)

Since the solution is neutral, the concentrations of H鈦� and OH鈦� ions are equal, and can be found using the ion product of water (Kw), which is equal to 1 脳 10鈦宦光伌. Kw = [H鈦篯 脳 [OH鈦籡 1 脳 10鈦宦光伌 = [H鈦篯 脳 [OH鈦籡 Since [H鈦篯 = [OH鈦籡, we can write: 1 脳 10鈦宦光伌 = [H鈦篯虏 [H鈦篯 = 鈭�(1 脳 10鈦宦光伌) = 1 脳 10鈦烩伔 Therefore, for 1.0 M KCl solution: [OH鈦籡 = 1 脳 10鈦烩伔 M [H鈦篯 = 1 脳 10鈦烩伔 M And the pH can be calculated using the formula: pH = -log鈧佲個[H鈦篯 pH = -log鈧佲個(1 脳 10鈦烩伔) = 7 So the pH of 1.0 M KCl solution is 7.
03

Calculate [OH鈦籡, [H鈦篯, and pH for a basic solution (1.0 M KC鈧侶鈧僌鈧�)

Since the solution is basic, we need to determine the initial concentration of the weak acid formed from the salt, which is CH鈧僀OOH. The initial concentration of CH鈧僀OOH is equal to the concentration of the salt KC鈧侶鈧僌鈧�, which is 1.0 M. The equilibrium reaction for the ionization of CH鈧僀OOH is: CH鈧僀OOH + H鈧侽 鈬� CH鈧僀OO鈦� + H鈧僌鈦� (H鈧僌鈦� is the hydrated form of H鈦�) We will assume that the ionization of CH鈧僀OOH is small and can be approximated by x. Therefore, at equilibrium: [CH鈧僀OO鈦籡 = x [H鈦篯 = x [CH鈧僀OOH] = 1 - x 鈮� 1 (because x << 1) Now we can use the Ka expression for CH鈧僀OOH to relate x and the concentrations: Ka = ([CH鈧僀OO鈦籡 脳 [H鈦篯) / [CH鈧僀OOH] 1.8 脳 10鈦烩伒 (given Ka for CH鈧僀OOH) = (x 脳 x) / 1 x虏 = 1.8 脳 10鈦烩伒 x = 鈭�(1.8 脳 10鈦烩伒) x = [H鈦篯 鈮� 4.24 脳 10鈦宦� M Now we can calculate the pH using the formula: pH = -log鈧佲個[H鈦篯 pH 鈮� -log鈧佲個(4.24 脳 10鈦宦�) 鈮� 2.37 To find [OH鈦籡, we use the Kw expression again: Kw = [H鈦篯 脳 [OH鈦籡 [OH鈦籡 = Kw / [H鈦篯 [OH鈦籡 = (1 脳 10鈦宦光伌) / (4.24 脳 10鈦宦�) 鈮� 2.36 脳 10鈦宦孤� M So for 1.0 M KC鈧侶鈧僌鈧�: [OH鈦籡 鈮� 2.36 脳 10鈦宦孤� M [H鈦篯 鈮� 4.24 脳 10鈦宦� M pH 鈮� 2.37

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Chemistry
Acid-base chemistry is a fundamental topic that explores the behavior of acids and bases in solution. Acids are substances that can donate protons (H鈦� ions), while bases are substances that can accept protons.
In the context of substance descriptions, we often categorize solutions based on their pH values:
  • Acidic solutions have a pH less than 7.
  • Neutral solutions have a pH exactly equal to 7.
  • Basic (or alkaline) solutions have a pH greater than 7.
Understanding the nature of a solution helps predict chemical reactions and how the solution will behave under different conditions.
In the exercise, determining whether KCl and KC鈧侶鈧僌鈧� are acidic, basic, or neutral sets the stage for further calculations.
Ion Product of Water
The ion product of water, or Kw, is a critical constant in acid-base chemistry that expresses the relationship between the concentration of hydrogen ions \(\text{H}^+\) and hydroxide ions \(\text{OH}^-\) in water.
Mathematically, this relationship is represented by: \[ K_w = [\text{H}^+] \times [\text{OH}^-] = 1 \times 10^{-14} \] at 25掳C. This means that the product of the concentrations of these ions in pure water or a neutral solution is always \(1 \times 10^{-14}\).
This concept is pivotal when determining the pH of solutions, especially neutral ones like 1.0 M KCl, where \(\text{H}^+\) and \(\text{OH}^-\) are equal, thus both equal to \(1 \times 10^{-7} \) M.
Equilibrium Reactions
Equilibrium reactions are reactions that occur in a closed system where reactants and products are present at concentrations that do not change over time.
In the exercise, when dealing with KC鈧侶鈧僌鈧�, we consider the equilibrium reaction of acetic acid ionization: \[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \] Here, acetic acid (CH鈧僀OOH) partially ionizes in water, setting up an equilibrium between the acid and its ionization products.
The ionization extent is determined by its acid dissociation constant (\(K_a\)), which gives us an insight into the strength of the weak acid and helps us calculate the equilibrium concentrations of ions.
Strong Acids and Bases
Strong acids and bases are substances that dissociate completely in water, meaning they donate or accept protons readily and leave no undissociated molecules.
When KCl is dissolved in water, it results from the combination of a strong acid (HCl) and a strong base (KOH), both of which do not remain in their undissociated forms, hence forming a neutral solution.
On the other hand, salts like KC鈧侶鈧僌鈧� arise from a strong base (KOH) and a weak acid (CH鈧僀OOH). This characteristic makes the solution basic, as the strong base dominates.
  • Strong acids include HCl, H鈧係O鈧�, and HNO鈧�.
  • Strong bases include KOH, NaOH, and LiOH.
Understanding which components of a salt are strong or weak indicates whether the resulting solution will be acidic, basic, or neutral.

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