Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4A) 3.09B) 4.11C) 3.82D) 3.46E) 2.78

Solution: A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4. A) 3.09B) 4.11C) 3.82D) 3.46E)

Problem

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4

A) 3.09

B) 4.11

C) 3.82

D) 3.46

E) 2.78

Solution

We’re being asked to calculate the pH of a buffer solution after the addition of HCl solution. The buffer solution is made up of 0.250 M HF and 0.250 M NaF.

Since the solution is a buffer, we know that HF is a weak acid and NaF is the conjugate base. HCl will react with the base NaF.

Reaction:         NaF(aq) + HCl(aq) → HF(aq) + NaCl(aq)


We will calculate the pH of the solution using the following steps:

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