Acid – Base Math Problems
Basic Equations:
[H+(aq)][OH-(aq)] = Kw = 1.0 x 10-14
since: Kw = [H+(aq)][OH-(aq)]
then: [H+(aq)] = Kw__
[OH-(aq)]
and: [OH-(aq)] = Kw__
[H+(aq)]
pH = -log[H+(aq)] and [H+(aq)] = 10-pH
pOH = -log[OH-(aq)] and [OH-(aq)] = 10-pOH
pH + pOH = 14
pOH = 14 – pH
pH = 14 – pOH
p = concentration of acid ionized x 100%
concentration of acid solute
where p = percent ionization
p = [H+(aq)] x 100% or [H+(aq)] = p x [HA(aq)]
[HA(aq)]
100
where [HA(aq)] = conc. of the acid
Ka = acid ionization constant Ka = [H+(aq)][A-(aq)]
[HA(aq)]
for acetic acid; HC2H3O2(aq)⇔H+(aq) + HC2H3O2-(aq)
Ka = [H+(aq)][HC2H3O2-(aq)]
[HC2H3O2(aq)]
Kb = base ionization constant Kb = [HB-(aq)][OH-(aq)]
[B(aq)]
Where
HA = acid A- = (anion) conjugate base
HB = conjugate base B = base
for ammonia; NH3(aq) + H2O(aq)⇔OH-(aq) + NH4+(aq)
Kb = [OH-(aq)][NH4+(aq)]
[NH3(aq)]
Kw = KaKb Ka = Kw
Kb
Kb = Kw
Ka
See Appendix C9 for Ka and Kb values
1) Calculating [H+(aq)] or [OH-(aq)] for a strong acid or
strong base – and then calculating pH or pOH
Sample Questions:
Calculate the concentration of the hydroxide ions of a 0.15 mol/L solution of
hydrochloric acid at SATP.
Since HCl is a strong acid, we assume it undergoes 100% ionization in aqueous
solution. HCl(g)⇔ H+(aq) + Cl-(aq) ∴ [HCl] = [H+(aq)][OH-(aq)]
Water is a weak electrolyte and ionizes very little: H2O(l)⇔H+(aq) + OH-(aq);
In pure water, [H+(aq)] = [OH-(aq)] = 1.0 x 10-7 mol/L.
Kw = 1.0 x 10-14
However as HCl is added, the added H+(aq) will shift equilibrium (in the equation
above) to the left according to Le Châtelier’s principle.
The concentration of H+(aq) from HCl is very large, so the contribution of H+(aq) from
the autoionization of water is insignificant.
[H+(aq)] is 0.15 mol/L.
[OH-(aq)] =
Kw = [H+(aq)][OH-(aq)]= 1.0 x 10-14
Kw__= 1.0 x 10-14 = 6.7 x 10-14 mol/L
[H+(aq)]
0.15
pOH = -log[OH-(aq)]
pH = 14 – pOH
pOH = -log(6.7 x 10-14 mol/L) = 13.2
pH = 14 – 13.2 = 0.8
Calculate the hydrogen ion concentration in a 0.25 mol/L solution of barium
hydroxide (a strong base).
Ba(OH)2(aq)⇔ Ba2+(aq) + 2OH-(aq)
If [Ba(OH)2(aq)] = 0.25 mol/L then [OH-(aq)] = 2 x 0.25 mol/L = 0.50 mol/l
[OH- (aq)] is 0.50 mol/L.
Kw = [H+(aq)][OH-(aq)]= 1.00 x 10-14
[H+(aq)] = Kw__
[H+(aq)] =1.00 x 10-14 = 2.00 x 10-14
[OH-(aq)]
0.50
pH = -log[H+(aq)] = - log(2.00 x 10-14) = 13.7
2) Determining the Percent Ionization of Weak Acids
Sample Question:
The pH of a 0.10 mol/L methanoic (formic) acid solution is 2.38. Calculate the
percent ionization of methanoic acid.
HCO2H(aq) + H2O(l) ↔ HCO2-(aq) + H3O+(aq)
Or more easily stated:
HCO2H(aq) ↔ HCO2-(aq) + H+(aq)
pH = 2.38,
therefore [H+(aq)] = 10-pH
[H+(aq)] = 10-2.38
[H+(aq)] = 4.2 x 10-3 mol/L
[H+(aq)] = p x [HCO2H(aq)] [HCO2H(aq)] = concentration of the methanoic acid
100
p=
[H+(aq)] x 100%
p = 4.2 x 10-3 mol/L x 100% = 4.2%
[HCO2H(aq)]
0.1 mol/L
3) Determine the pH of a Weak Acid given Percent
Ionization
1.3%
HC2H3O2(aq) + H2O(l) ↔ C2H3O2-(aq) + H3O+(aq)
p = concentration of acid ionized x 100%
concentration of acid solute
p = [H+(aq)] x 100%
[HA(aq)]
or
[H+(aq)] = p x [HA(aq)] [HA(aq)] = conc. of the acid
100
Determine the pH of 0.1 mol/L
Acetic Acid (HC2H3O2(aq)) with a
percent ionization of 1.3%
Since pH = - log [H+(aq)]
p = percent ionization
[H+(aq)] = p x [HA(aq)]
100
+
[H (aq)] = 1.3 x [0.1 mol/L]
100
+
[H (aq)] = 1.3 x 10-3 mol/L
pH = -log(1.3 x 10-3 mol/L)
= 3.75
4) Calculate Ka from percent ionization
Calculate the acid ionization constant, Ka, of acetic acid if a 0.1000 mol/L
solution at equilibrium at SATP has a percent ionization of 1.3%
HC2H3O2(aq)↔ H+(aq)
+
C2H3O2-(aq)
Ka = [H+(aq)][C2H3O2-(aq)]
[HC2H3O2(aq)]
I
C
E
HC2H3O2(aq)
0.1000
-x
0.1000 - x
I
C
E
HC2H3O2(aq)
0.1000
- 0.0013
0.0987
Ka = [H+(aq)][C2H3O2-(aq)]
[HC2H3O2(aq)]
since [HC2H3O2(aq)] = 0.1000 mol/L
and 1.3% of the acid dissociates
Therefore [H+(aq)] and [C2H3O2-(aq)] = 0.1000 x 0.013
= 0.0013 mol/L
+
H (aq)
C2H3O2-(aq)
0
0
+x
+x
+x
+x
H+(aq)
0
- 0.0013
0.0013
C2H3O2-(aq)
0
+ 0.0013
0.0013
Ka = (0.0013)(0.0013) = 1.7 x 10-5
(0.0987)
5) Calculate Ka given Kb (or calculate Kb given Ka)
What is the value of the base ionization constant, Kb for the acetate ion,C2H3O2(aq), at SATP?
Kw = 1.0 x 10-14 Ka = 1.8 x 10-5
Kw = KaKb
Kb = Kw = 1.0 x 10-14 = 5.6 x 10-10
Ka1.8 x 10-5
6) Calculate [H+(aq)] and pH of a Weak Acid, given Ka
Calculate [H+(aq)] and pH of 0.10 mol/L acetic acid solution. Ka for acetic acid is 1.8
x 10-5.
HC2H3O2(aq)↔ H+(aq)
+
C2H3O2-(aq)
Ka = [H+(aq)][C2H3O2-(aq)]
[HC2H3O2(aq)]
I
C
E
Ka =
HC2H3O2(aq)
0.10
-x
0.10 - x
(x)(x) = 1.8 x 10-5
(0.10 – x)
Ka = 1.8 x 10-5
[HC2H3O2(aq)] = 0.10 mol/L
H+(aq)
0
+x
+x
C2H3O2-(aq)
0
+x
+x
x2 = 1.8 x 10-5 (0.10 – x) ; but x is very small
x2 ≈ 1.8 x 10-5 (0.10)
x2 ≈ 1.8 x 10-6
x ≈ 1.3 x 10-3 (this value = [H+(aq)])
[HC2H3O2(aq)] = 0.10 – 1.3 x 10-3 = 0.0987 (not required for answer)
[H+(aq)] = 1.3 x 10-3
pH = -log[H+(aq)]
= -log(1.3 x 10-3)
= 2.89
7) Find Ka, given concentration and pH
The pH of a 0.100 mol/L hypochlorous acid (HClO) solution is 4.23. What is the Ka
for hypochlorous acid?
HClO(aq)↔ H+(aq)
+
ClO-(aq)
Ka = [H+(aq)][ClO-(aq)]
[HClO(aq)]
[HClO(aq)] = 0.100 mol/L; [H+(aq)] = 10-pH
[H+(aq)] = 10-pH
[H+(aq)] = 10-4.23
[H+(aq)] = 5.9 x 10-5 mol/L
[H+(aq)] = [ClO-(aq)] = 5.9 x 10-5 mol/L
Ka = [H+(aq)][ClO-(aq)] = (5.9 x 10-5 mol/L)(5.9 x 10-5 mol/L) = 3.5 x 10-8
[HClO(aq)]
0.100 mol/L
Practice page 570 #9-10
8) Calculate the [H+] and pH of a weak base given Ka
or Kb
 Kb values for bases are found in appendix C7 on page 803.
 However some Kb values are not listed. To find these Kb values, use Appendix
C9 on page 803 to find the Ka value of its conjugate acid.
 Use the formula, KaKb = Kw or Kb = Kw/Ka to determine Kb
 Solving for Kb is similar to solving for Ka
Calculate the pH of a 0.100 mol/L solution of hydrazine, N2H4, a weak base.
N2H4(aq) + H2O(l)↔ N2H5+(aq) + OH-(aq)
Kb = 1.7 x 10-6
H2O(l) ↔ OH-(aq) + H+(aq)
Kw = 1 x 10-14
Since Kw <<Kb the contribution of OH-(aq) by water is very small compared to N2H4
Kb = [N2H5+(aq)][ OH-(aq)]
[N2H4(aq)][H2O(l)]
I
C
E
Ka =
N2H4(aq)
0.100
-x
0.100 - x
(x)(x) = 1.7 x 10-6
(0.100 – x)
Kb = [N2H4(aq)][OH-(aq)] Kb = 1.7 x 10-6
[N2H4(aq)]
N2H4(aq)
0
+x
+x
OH-(aq)
0
+x
+x
x2 = 1.7 x 10-6 (0.10 – x) ; but x is very small
x2 ≈ 1.7 x 10-6 (0.10)
x2 ≈ 1.7 x 10-7
x ≈ 4.12 x 10-4 (this value = [OH-(aq)])
[N2H4(aq)] = 0.100 – 4.12 x 10-4 = 0.0996 (not required for answer)
[OH-(aq)] = 4.12 x 10-4
pOH = -log[OH-(aq)]
pH = 14.0 - pOH
-4
= -log(4.12 x 10 )
= 10.62
= 3.38
Practice page 574 #12-13
9) Determining the pH of Polyprotic Acids
 Polyprotic acids are acids with more than one H+(aq) to release into solution
 For polyprotic acids the first ionization (the first H+(aq) released) is known as
Ka1 and the second ionization as Ka2.
 Since Ka2 is much smaller than Ka1, Ka1 determines the pH of the solution.
Calculate the pH of a 0.10 mol/L solution of ascorbic acid, H2C6H6O6.
Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12
H2C6H6O6 ↔ H+ + HC6H6O6Ka = [H+(aq)][HC6H6O6-(aq)]
[H2C6H6O6(aq)]
I
C
E
Ka =
H2C6H6O6(aq)
0.10
-x
0.10 - x
(x)(x) = 7.9 x 10-5
(0.10 – x)
Ka1 = 7.9 x 10-5
Ka = 1.8 x 10-5
[H2C6H6O6(aq)] = 0.10 mol/L
H+(aq)
0
+x
+x
HC6H6O6-(aq)
0
+x
+x
x2 = 7.9 x 10-5 (0.10 – x) ; but x is very small
x2 ≈ 7.9 x 10-5 (0.10)
x2 ≈ 7.9 x 10-6
x ≈ 2.8 x 10-3 (this value = [H+(aq)])
[H+(aq)] = 2.8 x 10-3
pH = -log[H+(aq)]
= -log(2.8 x 10-3)
= 2.55
The HC6H6O6- can ionize a second time to produce more H+ ions and decrease the
pH more:
HC6H6O6- ↔ H+ + C6H6O62Ka2 = 1.6 x 10-12
However since the HC6H6O6- ion is a much weaker acid than H2C6H6O6 (compare Ka
values: Ka1/Ka2 = 7.9 x 10-5/1.6 x 10-12 = 4.9 x 10-7, the Ka2 value will have an
insignificant effect on the pH. As a result the pH is still 2.55
Practice page 578 # 14
Summative Assignment
page 579 # 1-4, 6, 7a, 9, 11-12, 16
10) Determining the pH of Salt Solutions
Calculate the pH of a 0.10 mol/L solution of NaNO2(aq)
NaNO2(aq) ↔ Na+ + NO2-.
Na+ is the very weak conjugate acid of a strong base
NO2- is the weak conjugate base of a weak acid
hence, NaNO2(aq) will form a basic solution
since NO2-(aq) is the base it reacts with water:
NO2-(aq) + H2O(l) ↔ HNO2(aq) + OH-(aq) Kb = ?
We know that for HNO2(aq), Ka = 7.2 x 10-4. NO2-(aq) is its conjugate base, so:
Kw = KaKb Kb = Kw
Ka = 1.0 x 10-14 = 1.4 x 10-11.
Ka
7.2 x 10-4
Kb = [HNO2(aq)][OH-(aq)]
[NO2-(aq)][H2O(l)]
Kb = [HNO2(aq)][OH-(aq)]
[NO2-(aq)]
Kb = 1.4 x 10-11
I
C
E
Ka =
NO2- (aq)
0.100
-x
0.10 - x
(x)(x) = 1.4 x 10-11
(0.100 – x)
OH-(aq)
0
+x
+x
HNO2(aq)
0
+x
+x
x2 = 1.4 x 10-11 (0.10 – x) ; but x is very small
x2 ≈ 1.4 x 10-11 (0.10)
x2 ≈ 1.4 x 10-12
x ≈ 1.18 x 10-6 (this value = [OH-(aq)])
[OH-(aq)] = 1.18 x 10-6
pOH = -log[OH-(aq)]
= -log(1.18 x 10-6)
= 5.92
pH = 14.0 - pOH
= 8.08
Calculate the pH of a 0.20 mol/L solution of NH4Cl(aq)
NH4Cl(aq) ↔ NH4+(aq) + Cl-(aq)
NH4+(aq) is the weak conjugate acid of a weak base
Cl- is the very weak conjugate base of a strong acid
hence, NH4Cl(aq) will form an acidic solution
since NH4+(aq) is the acid it reacts with water:
NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq) Ka = ?
OR: NH4+(aq) ↔ NH3(aq) + H+(aq) Ka = ?
We know that for NH3(aq), Kb = 1.8 x 10-5. NH4+(aq) is its conjugate acid, so:
Kw = KaKb Ka = Kw
Kb = 1.0 x 10-14 = 5.8 x 10-10.
Kb
1.8 x 10-5
Ka = [NH4+(aq)][H3O+(aq)]
[NH3(aq)][H2O(l)]
Ka = [NH4+(aq)][H+(aq)] Ka = 5.8 x 10-10
[NH3(aq)]
I
C
E
Ka =
NH3 (aq)
0.20
-x
0.20 - x
(x)(x) = 5.8 x 10-10
(0.20 – x)
NH4+(aq)
0
+x
+x
x2 = 5.8 x 10-10 (0.20 – x) ; but x is very small
x2 ≈ 5.8 x 10-10 (0.20)
x2 ≈ 2.9 x 10-9
x ≈ 5.4 x 10-5 (this value = [H+(aq)])
[H+(aq)] = 5.4 x 10-5
H+(aq)
0
+x
+x
pH = -log[H+(aq)]
= -log(5.4 x 10-5)
= 4.2