Bronsted Lowry Theory Examples
Example 1: Calculate the pH of a solution formed by mixing 50 mL of 0.1 M hydrochloric acid (HCl) with 150 mL of 0.05 M sodium hydroxide (NaOH). Given that the dissociation constants (Ka) of HCl and NaOH are 1.0 × 106 and 1.0 × 10-14 respectively.
Solution:
Step 1: Determine the number of moles of HCl and NaOH
Moles of HCl = 50 mL × 0.1 mol/L = 0.005 mol
Moles of NaOH = 150 mL × 0.05 mol/L = 0.0075 mol
Step 2: Identify the limiting reagent (reactant that gets used up first)
HCl is the limiting reagent (0.005 mol vs. 0.0075 mol)
Step 3: Calculate the excess moles of NaOH after the reaction with HCl
Excess moles of NaOH = Moles of NaOH – Moles reacted with HCl
= 0.0075 mol – 0.005 mol = 0.0025 mol
Step 4: Calculate the concentration of OH- ions from the excess NaOH.
Concentration of OH- ions = Excess moles of NaOH / Total volume of the solution
= 0.0025 mol / (50 mL + 150 mL)
= 0.0025 mol / 200 mL
= 0.0025 mol / 0.2 L
= 0.0125 mol/L
Step 5: Calculate pOH using the concentration of OH- ions
pOH = -log(OH- concentration) = -log(0.0125) ≈ 1.9
Step 6: Calculate pH using the relation pH + pOH = 14 (for water at 25°C)
pH = 14 – pOH ≈ 14 – 1.9 ≈ 12.1
Therefore, the pH of the solution ≈ 12.1.
Example 2: What volume of 0.2 M acetic acid (CH₃COOH) must be added to 100 mL of water to create a solution with a pH of 4.0? (Ka for acetic acid is 1.8 × 10-5).
Solution:
Step 1: Calculate the concentration of H+ ions corresponding to pH 4.0
[H+] = 10(-pH) = 10(-4.0) = 1.0 × 10(-4) mol/L
Step 2: Use the given Ka value to calculate the concentration of CH₃COOH
Ka = [H+][CH₃COO-] / [CH₃COOH]
Given [H+] = [CH₃COO-] (for a weak acid)
[CH₃COOH] = [H+] × [CH₃COOH] / Ka
= (1.0 × 10(-4)) × [CH₃COOH] / (1.8 × 10(-5))
[CH₃COOH] = 5.56 × 10-1 mol/L
Step 3: Use the formula for concentration (C = n/V) to find the volume of CH₃COOH needed.
Volume = n / C
Volume = 5.56 × 10-1 mol / 0.2 mol/L ≈ 2.78 L
Therefore, approximately 2.78 liters of 0.2 M acetic acid need to be added.
Example 3: Determine the pH of a solution obtained by mixing 25 mL of 0.1 M hydrochloric acid (HCl) with 75 mL of 0.2 M sodium acetate (CH₃COONa). (Ka for Acetic Acid is 1.8 × 10-5).
Solution:
Step 1: Identify the components that will react.
HCl will fully dissociate and contribute H+ ions.
Sodium acetate (CH3COONa) will partially dissociate into acetate ions (CH3COO–) and Na+ ions.
Step 2: Calculate the moles of H+ ions from HCl and CH3COO– ions from sodium acetate.
Moles of H+ ions from HCl = 25 mL × 0.1 mol/L = 0.0025 mol
Moles of CH3COO– ions from CH3COONa = 75 mL × 0.2 mol/L = 0.015 mol
Step 3: Calculate the total moles of CH3COO– ions.
Total moles of CH3COO– ions = Moles from sodium acetate – Moles reacted with HCl
= 0.015 mol – 0.0025 mol = 0.0125 mol
Step 4: Calculate the concentration of CH3COO– ions.
Total Volume of Solution = 25 mL + 75 mL = 100 mL = 0.1 L
Concentration of CH3COO– ions = Moles / Volume
= 0.0125 mol / 0.1 L = 0.125 mol/L
Step 5: Calculate pOH using the concentration of CH3COO– ions and Ka
pOH = -log(CH3COO– concentration) = -log(0.125) ≈ 0.9
Step 6: Calculate pH using the relation pH + pOH = 14 (for water at 25°C)
pH = 14 – pOH ≈ 14 – 0.9 ≈ 13.1
Therefore, the pH of the solution ≈ 13.1
Example 4. A 0.05 M solution of formic acid (HCOOH) has a pH of 2.6. Calculate the concentration of hydroxide ions ([OH-]) in this solution. (Ka for formic acid is 1.8 × 10-4).
Solution:
Given,
- pH = 2.6
pH = -log[H+]
[H+] = 10-pH
[H+] = 10(-2.6) ≈ 2.51 × 10-3mol/L
For a weak acid (HCOOH), use the relation Ka = [H+][HCOO–] / [HCOOH]
Given [H+] = [HCOO–] (for a weak acid)
Ka = [H+]2 / [HCOOH]
[HCOOH] = [H+]2 / Ka
[HCOOH] = (2.51 × 10(-3))2 / (1.8 × 10^(-4)) ≈ 3.54 × 10(-2) mol/L
Concentration of Hydroxide Ions is given by [OH–] = Kw / [H+].
Kw = 1.0 × 10-14 (at 25°C)
[OH–] = Kw / [H+] = 1.0 × 10^(-14) / 2.51 × 10(-3) ≈ 3.98 × 10(-12) mol/L
Therefore, the concentration of hydroxide ions ≈ 3.98 × 10^(-12) mol/L.
Example 5: Calculate the pH of a solution prepared by mixing 25 mL of 0.1 M hydrochloric acid (HCl) and 75 mL of 0.05 M sodium hydroxide (NaOH). (Ka for HCl is 1.0 × 106 and Kw for water is 1.0 × 10-14).
Solution:
Step 1: Determine the moles of H+ ions from HCl and OH- ions from NaOH.
Moles of H+ ions from HCl = 25 mL × 0.1 mol/L = 0.0025 mol
Moles of OH– ions from NaOH = 75 mL × 0.05 mol/L = 0.00375 mol
Step 2: Identify the limiting reagent (HCl or NaOH) based on the available moles.
HCl is the limiting reagent (0.0025 mol vs. 0.00375 mol)
Step 3: Calculate the excess moles of OH– ions after the reaction with HCl
Excess moles of NaOH = Moles of NaOH – Moles reacted with HCl
= 0.00375 mol – 0.0025 mol = 0.00125 mol
Step 4: Calculate the concentration of OH- ions from the excess NaOH
Concentration of OH– ions = Excess moles of NaOH / Total volume of the solution
= 0.00125 mol / (25 mL + 75 mL)
= 0.00125 mol / 0.1 L
= 0.0125 mol/L
Step 5: Calculate pOH using the concentration of OH– ions.
pOH = -log(OH- concentration) = -log(0.0125) ≈ 1.9
Step 6: Calculate pH using the relation pH + pOH = 14 (for water at 25°C)
pH = 14 – pOH ≈ 14 – 1.9 ≈ 12.1
Therefore, the pH of the solution ≈ 12.1
What Is Bronsted–Lowry Theory?
Bronsted-Lowry Theory, also called the Proton Theory of Acid and Base, is a theory that explains the concept of acid and base. It was given by Johannes Nicolaus Bronsted (Danish Chemist) and Thomas Martin Lowry (English Chemist) in 1923.
In this article, we will learn about, Bronsted Lowry’s Theory Definition, Examples, and others in detail.
Table of Content
- What is Bronsted Lowry Theory?
- Bronsted-Lowry Acids and Base
- Advantages of Bronsted Lowry Theory of Acid and Base
- Disadvantages of Bronsted Lowry Theory
- Applications of Bronsted Lowry Theory of Acid and Base
- Difference between Arrhenius Theory, Bronsted-Lowry Theory, and Lewis Acid-Base Theory