SME
General Chemistry II
Acid–base equilibria and buffers
Use this for “Find the pH and percent ionization of 0.150 M nitrous acid, HNO₂ (Kₐ = 4.5 × 10⁻⁴).”
Diagram
Since pH = −log₁₀[H⁺], here [H⁺] = 10⁻²·¹⁰ = 0.007994 mol/L.
Below 7: acidic.
This is an example. Type your own values to start a new problem.
Kₐ
=
(
[H⁺] (M)
)
2
÷
(
C (M)
−
[H⁺] (M)
)
pH
=
−log(
[H⁺] (M)
)
=
C is the acid’s concentration before any of it ionizes, and Kₐ its acid dissociation constant.
pKₐ
Acid strength as a log
calculated
%
Percent ionized
calculated · %
Formulas
Kₐ
=
[H⁺]²
÷
(
C
−
[H⁺])
4.5 ×
10
-4
=
0.00799
2
0.15 − 0.00799
pH = −log₁₀([H⁺])
2.10 = −log₁₀(0.00799)
pKₐ = −log₁₀(Kₐ)
3.35
=
−log₁₀(4.5
×
10
-4
)
%
=
100
×
[H⁺]
÷
C
5.33
=
100
×
0.00799
0.15
Assumptions
•
A weak acid HA ⇌ H⁺ + A⁻. If x is the [H⁺] it makes, then [A⁻] = x and [HA] = C − x at equilibrium.
•
Kₐ = x² ÷ (C − x) is solved exactly by the quadratic formula. The shortcut x ≈ √(KₐC) is fine when it lands within 5%.
•
Water’s own H⁺ is left out, which is fine while x is above 10⁻⁶ M.
•
The water is at 25 °C. A pH has as many decimals as the concentration has significant figures.
Step-by-step
WE KNOW
Kₐ = 4.5 × 10⁻⁴, C = 0.15 M
FIND
hydrogen ion concentration ([H⁺]), pH (pH), acid strength as a log (pKₐ), percent ionized (%)
Step 1 · Find hydrogen ion concentration ([H⁺])
Use Kₐ = [H⁺]² ÷ (C − [H⁺])
Kₐ
=
x
2
C
− x
is
x
2
+
Kₐx
−
KₐC
=
0.
Its
positive
root
is
−Kₐ +
√
Kₐ² + 4KₐC
2,
written
here
as
2Kₐ
C
Kₐ +
√
Kₐ² + 4KₐC
so
a
tiny
x
keeps
its
digits.
[H⁺]
=
2
×
Kₐ
×
C
Kₐ +
√
Kₐ² + 4 × Kₐ ×
C
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
÷
(4.5
×
10
-4
+
√(
(4.5 ×
10
-4
)
2
+
4
×
(4.5
×
10
-4
)
×
0.15))
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
÷
((4.5
×
10
-4
)
+
√((2.025
×
10
-7
)
+
4
×
(4.5
×
10
-4
)
×
0.15))
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
÷
((4.5
×
10
-4
)
+
√((2.025
×
10
-7
)
+
0.0018
×
0.15))
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
÷
((4.5
×
10
-4
)
+
√((2.025
×
10
-7
)
+
0.00027))
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
(4.5 ×
10
-4
) +
√
0.0002702
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
(4.5 ×
10
-4
) + 0.01644
[H⁺]
=
2
×
(4.5
×
10
-4
)
×
0.15
0.01689
[H⁺]
=
0.0009
×
0.15
0.01689
[H⁺]
=
0.000135
0.01689
[H⁺]
=
0.00799
M
→
The
shortcut
√
KₐC
gives
0.00822
M,
2.8%
off:
close
enough
to
use
Step 2 · Find pH (pH)
Use pH = −log₁₀([H⁺])
pH counts the powers of ten in [H⁺].
pH = −log₁₀(0.00799)
pH = 2.10
Step 3 · Find acid strength as a log (pKₐ)
Use pKₐ = −log₁₀(Kₐ)
pKₐ puts Kₐ on the pH scale: the smaller the pKₐ, the stronger the acid.
pKₐ
=
−log₁₀(4.5
×
10
-4
)
pKₐ = 3.35
Step 4 · Find percent ionized (%)
Use % = 100 × [H⁺] ÷ C
The share of the acid that came apart, as a percent.
%
=
100
×
0.00799
0.15
%
=
0.799
0.15
% = 5.33%
Check
Put the numbers back in. Both sides should match.
4.5 ×
10
-4
=
0.00799
2
0.15 − 0.00799
✓
2.10 = −log₁₀(0.00799) ✓
3.35
=
−log₁₀(4.5
×
10
-4
)
✓
5.33
=
100
×
0.00799
0.15
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
pH of a weak base
Use this for “Find the pOH and pH of 0.250 M methylamine, CH₃NH₂ (Kb = 4.4 × 10⁻⁴).”
Buffer pH
Use this for “A buffer holds 0.200 mol HF and 0.250 mol F⁻ (pKₐ = 3.17). Find its pH, then its pH after adding 0.020 mol NaOH.”
pH of a salt solution
Use this for “Find the pH of 0.250 M sodium fluoride, NaF. Kₐ of HF is 6.8 × 10⁻⁴.”
pH at the equivalence point
Use this for “20.00 mL of 0.120 M formic acid (Kₐ = 1.8 × 10⁻⁴) is titrated with 0.150 M NaOH. Find the pH at the equivalence point.”