Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
All Chapters
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
Problem

Calculate the pH of a 0.10 M solution of sodium formate (NaHCOO) given that the Ka of formic acid (HCOOH) is 1.8 x 10-4.

Solution

We are being asked to calculate the pH for a 0.10 M solution of C6H5NH2. 


NaHCOO is the salt form of the conjugate base of HCOOH.


NaHCOO disociates as: 

NaHCOO(aq) → Na+(aq) + HCOO-(aq)


The acid dissociation goes as: 

HCOOH(aq) + H2O(l) → HCOO-(aq) + H3O+(aq)



Since we’re dealing with a weak base and Kb is an equilibrium expression we will have to create an ICE chart to determine the equilibrium concentration of each species:

HCOO-(aq) weak baseproton acceptor
H2O → will act as the weak acidproton donor


Equilibrium reaction:        HCOO-(aq)  + H2O(l)  HCOOH(aq) + OH-(aq)


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