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
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: Determine the pH of a solution that is 3.65% KOH by mass. Assume that the solution has density of 1.01  g / mL.

Solution: Determine the pH of a solution that is 3.65% KOH by mass. Assume that the solution has density of 1.01  g / mL.

Problem

Determine the pH of a solution that is 3.65% KOH by mass. Assume that the solution has density of 1.01  g / mL.

Solution

We’re being asked to calculate the pH of a solution that is of 3.65% KOH by mass. Assume that the solution has density of 1.01 g/mL. 


KOH is a strong base.

Remember that strong bases completely dissociate in water and that bases accept H+ from the acid (water in this case). 


The dissociation of KOH is as follows:


KOH (aq)  K+ (aq) + OH-(aq)


For strong base, an ICE table is not necessary because it can be assumed that all of the acid or base will dissociate. 


The concentration of the strong base is the same as its dissociated particles.


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