Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
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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: Formic acid, Ka = 1.8x10 –4, is the principle component in the venom in stinging ants. What is the molarity of a formic acid solution if 25.00mL of formic acid solution requires 29.80mL of 0.0567M solution of sodium hydroxide to reach the equivalence point?
  1. 0.0676M
  2. 0.0134M
  3. 0.0567M
  4. 0.0476M

Solution: Formic acid, Ka = 1.8x10 –4, is the principle component in the venom in stinging ants. What is the molarity of a formic acid solution if 25.00mL of formic acid solution requires 29.80mL of 0.0567M sol

Problem

Formic acid, Ka = 1.8x10 –4, is the principle component in the venom in stinging ants. What is the molarity of a formic acid solution if 25.00mL of formic acid solution requires 29.80mL of 0.0567M solution of sodium hydroxide to reach the equivalence point?

  1. 0.0676M
  2. 0.0134M
  3. 0.0567M
  4. 0.0476M
Solution

We’re being asked to calculate the molarity of formic acid if 25.00 mL of formic acid required 29.80 mL of 0.0567 M NaOH to reach the equivalence point. Recall that at the equivalence point of a titration:



Recall that moles = molarity × volume. This means:



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