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
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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: For phosphoric acid, H3PO4, the pKas are:pKa1 = 2.2pKa2 = 7.2pKa = 12.7a) Write the reaction associated with K a2.b) Write Ka2 in terms of the concentration of the reactants and products in the reacti


For phosphoric acid, H3PO4, the pKas are:

pKa1 = 2.2

pKa2 = 7.2

pKa = 12.7

a) Write the reaction associated with K a2.

b) Write Ka2 in terms of the concentration of the reactants and products in the reaction above?

c) To make a buffer that is close to neutral pH, how would you use some of the following. Be precise in terms of what you would use and how much of each.

0.001 M HCl, 1 M HCl, 10 M HCl

0.001 M H3PO4, 1 M H3PO4

0.001 M NH3, 1 M NH3


Solids: Fe, Na, NaCl, NaH, NaOH


We are given the pKa(s) of Phosphoric Acid, H3PO4, a triprotic acid which can contribute up to 3 protons [H+] into solution. Each pKa corresponds to one proton dissociation. 

Let us write the chemical equation for each proton dissociation

a. pKa1 = 2.2         H3PO4(aq)   +    H2O(l)    ⇌     H2PO4-(aq)    +   H3O+(aq)

b. pKa2 = 7.2         H2PO4-(aq)   +    H2O(l)    ⇌     HPO42-(aq)    +   H3O+(aq)

c. pka3 = 12.7        HPO42-(aq)   +    H2O(l)    ⇌     PO43-(aq)    +   H3O+(aq)

In each chemical reaction, notice that the proton, H+, is released into the solution, forming H3Oreducing one proton from the acid each time.

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