Monoprotic Acids and Bases
Chapter 9
Monoprotic Acids and Bases
Bronsted and Lowry definitions
Acids are proton donars
Bases are proton acceptors
HCl(g) + NH3(g) ( NH4+Cl-(s)
acid base salt
CH3OOH + CH3NH2 ↔ CH3OO- + CH3NH3+
Acetic acid methyl amine acetate ion methyl ammonium ion
acid base conjugate base conjugate acid
For aqueous solutions,
Acids react with water to yield H3O+
Bases react with water to yield OH-
acid
CH3OOH + H2O ↔ CH3OO- + H3O+
Acetic acid water acetate ion hydronium ion
acid base conjugate base conjugate acid
conjugate pairs
CH3OOH
CH3OO-
base
CH3NH2 + H2O ↔ CH3NH3+ + OH-
methyl amine water methyl ammonium ion hydroxide ion
base acid conjugate acid conjugate base
conjugate pairs
CH3NH2
CH3NH3+
Autoprotolysis of water
H2O + H2O ↔ H3O+ + OH-
Equilibrium constant is Kw
Kw = [H3O+][OH-] = 1.01*10-14 @ 25 C
For pure water (what is pure water?)
[H3O+] = [OH-] = 1.00*10-7 M
pH – a measure of the acidity of a solution
pH = -log [H3O+] or –log AH+ (ch. 8)
below 7 is acidic
above 7 is neutral
Strong acids and bases – complete dissociation in water
Example of strong acid
What is the pH of a 0.2 M HCl solution
HCl is a strong acid
HCl + H2O → Cl- + H3O+
The equilibrium constant for this reaction is large. There is virtually no HCl species remaining in solution. Therefore,
[H3O+] = 0.2 M and pH = -log (0.2) = 0.7
0.02 M solution → 1.7
sig. fig. rule: the number of significant figures in the concentration is equal to the number of significant figures after the decimal point in the pH.
Example of strong base
What is the pH of a 0.020 M Ca(OH)2 solution?
Ca(OH)2 is a strong base.
Ca(OH)2 → Ca2+ + 2OH-
The equilibrium constant for this reaction is large. There is virtually no Ca(OH)2(s) remains. Therefore,
[OH-] = 0.040 M, [H3O+] = 2.5*10-13 M, and pH = -log (0.040) = 12.60
sig. fig. rule: the number of significant figures in the concentration is equal to the number of significant figures after the decimal point in the pH.
However, in these examples we have made an assumption. The assumption that has implicitly been made is that the [H+] from the strong acid (or [OH-] from the strong base) is >>>>> the [H+] (or the [OH-]) from water hydrolysis. Certainly these assumptions have been true!!!
In the strong acid example,
[H+]sa = 0.1 M
Kw = [OH-][H3O+] = 1.01*10-14
[OH-] = 1.01*10-13 M
[H3O+]water = [H+]water = 1.01*10-13 M
but what happens if this assumption breaks down?
Example:
Calculate the pH of a 5.0*10-8 M HClO4 solution
HClO4 is a strong acid, and thus completely dissociates to give H3O+ and ClO4- ions.
Sources of H3O+
HClO4 + H2O → H3O+ + ClO4-
2H2O ( H3O+ + OH-
pertinent equations;
[H3O+] = [H3O+]HClO4 + [H3O+]H2O
[OH-] = [H3O+]H2O
1.01*10-14 = Kw = [H3O+] [OH-]
[H3O+]HClO4 = 5.0*10-8 M
let x = [H3O+]H2O
substitutions;
Kw = x(5.0*10-8 + x) = 1.01*10-14
Assumption 1: Is [H3O+]H2O ................
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