Ch.15 - Acid and Base EquilibriumWorksheetSee 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
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: Determine the pH of a 0.00598 M HClO 4 solution.A) 11.777B) 6.434C) 7.566D) 2.223E) 3.558

Solution: Determine the pH of a 0.00598 M HClO 4 solution.A) 11.777B) 6.434C) 7.566D) 2.223E) 3.558

Problem

Determine the pH of a 0.00598 M HClO 4 solution.

A) 11.777

B) 6.434

C) 7.566

D) 2.223

E) 3.558

Solution

Remember that HClO4 is a strong, monoprotic oxyacid. It’s a strong oxyacid since there are 3 more O than H in the formula. 


Recall that strong acids completely dissociate in water and that acids donate H+ to the base (water in this case). The dissociation of HClO4 is as follows:


HClO4(aq) + H2O(l)  H3O+(aq) + ClO4(aq)


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