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Acids and BasesChapter 16
How do we measure pH?
• For less accurate measurements, one can use– Litmus paper
• Turns blue above ~pH = 8
• Turns red below ~pH = 5
– An indicator
How do we measure pH?
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.
Strong acids
• You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
• These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.
• For the monoprotic strong acids,
[H3O+] = [acid].
Strong bases
• Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).
• Again, these substances dissociate completely in aqueous solution.
Dissociation constants
• For a generalized acid dissociation,
the equilibrium expression would be
• This equilibrium constant is called the acid-dissociation constant, Ka.
[H3O+] [A−][HA]
Kc =
HA(aq) + H2O(l) A−(aq) + H3O+(aq)
Dissociation constants
The greater the value of Ka, the stronger the acid.
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]
= 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50 0.00
-x +x
0.50 - x
0.00
+x
x x
Ka =x2
0.50 - x= 7.1 x 10-4
Ka x2
0.50= 7.1 x 10-4
0.50 – x 0.50Ka << 1
x2 = 3.55 x 10-4 x = 0.019 M
[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72
[HF] = 0.50 – x = 0.48 M16.5
When can I use the approximation?
0.50 – x 0.50Ka << 1
When x is less than 5% of the value from which it is subtracted.
x = 0.0190.019 M0.50 M
x 100% = 3.8%Less than 5%
Approximation ok.
What is the pH of a 0.05 M HF solution (at 250C)?
Ka x2
0.05= 7.1 x 10-4 x = 0.006 M
0.006 M0.05 M
x 100% = 12%More than 5%
Approximation not ok.
Must solve for x exactly using quadratic equation.
16.5
Solving weak acid ionization problems:
1. Identify the major species that can affect the pH.
• In most cases, you can ignore the autoionization of water.
• Ignore [OH-] because it is determined by [H+].
2. Use ICE to express the equilibrium concentrations in terms of single unknown x.
3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.
4. Calculate concentrations of all species and/or pH of the solution.
16.5
What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122 0.00
-x +x
0.122 - x
0.00
+x
x x
Ka =x2
0.122 - x= 5.7 x 10-4
Ka x2
0.122= 5.7 x 10-4
0.122 – x 0.122Ka << 1
x2 = 6.95 x 10-5 x = 0.0083 M
0.0083 M0.122 M
x 100% = 6.8%More than 5%
Approximation not ok.
16.5
Ka =x2
0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0
ax2 + bx + c =0-b ± b2 – 4ac
2ax =
x = 0.0081 x = - 0.0081
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122 0.00
-x +x
0.122 - x
0.00
+x
x x
[H+] = x = 0.0081 M pH = -log[H+] = 2.09
16.5
percent ionization = Ionized acid concentration at equilibrium
Initial concentration of acidx 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0
x 100% [HA]0 = initial concentration
16.5
Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
• We know that
[H3O+] [COO−][HCOOH]
Ka =
Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
• To calculate Ka, we need the equilibrium concentrations of all three things.
• We can find [H3O+], which is the same as [HCOO−], from the pH.
Calculating Ka from the pH
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = [H3O+]
4.2 10−3 = [H3O+] = [HCOO−]
Calculating Ka from pH
Now we can set up a table…
[HCOOH], M [H3O+], M [HCOO−], M
Initially 0.10 0 0
Change −4.2 10-3 +4.2 10-
3
+4.2 10−3
At Equilibrium
0.10 − 4.2 10−3
= 0.0958 = 0.10
4.2 10−3 4.2 10−3
Calculating Ka from pH
[4.2 10−3] [4.2 10−3][0.10]
Ka =
= 1.8 10−4
Polyprotic Acids
• Have more than one acidic proton.
• If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Weak Bases and Base Ionization Constants
Kb =[NH4
+][OH-][NH3]
Kb is the base ionization constant
Kb
weak basestrength
16.6
Solve weak base problems like weak acids except solve for [OH-] instead of [H+].
16.6
16.7
Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq)
A- (aq) + H2O (l) OH- (aq) + HA (aq)
Ka
Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka = Kw
Kb
Kb = Kw
Ka
