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
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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
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Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solu

Problem

A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.

Determine the molecular weight (in g/mol) of the acid.

A. 84.8 g/mol

B. 249.6 g/mol

C. 499.1 g/mol

D. 269.4 g/mol

E. 49.9 g/mol

Solution

We’re being asked to calculate the molar mass of a weak monoprotic acid



Recall that at the equivalence point of a titration:


moles acid=moles base


Also, recall that moles = molarity × volume

This means:


MVacid=MVbase


We're going to calculate for the molar mass of the weak monoprotic acid using the following steps:

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