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

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO can be found here.

(a) before addition of any KOH

(b) after addition of 25.0 mL of KOH

(c) after addition of 30.0 mL of KOH

(d) after addition of 50.0 mL of KOH

(e) after addition of 60.0 mL of KOH


The reaction between HClO and KOH is given by this balanced equation:

HClO (aq) + KOH (aq) à KClO (aq) + H2O (aq)

Since HClO has a Ka value of 3.5x10-8 which is less than 1, it is a weak acid. Thus, this problem tackles a weak acid – strong base titration.

From the balanced equation, a mole-to mole comparison shows that 1 mole of HClO is equal to 1 mole of KOH.

This means that at every stage of the titration, the number of moles of the base reacting to the acid can be calculated.

The pH of the solution can then be determined using the excess acid or base that is not consumed in the reaction.

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