SME
General Chemistry II · Problem type
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.”
Diagram
Since pH = −log₁₀[H⁺], here [H⁺] = 10⁻⁸·²⁸ = 5.2154 × 10⁻⁹ mol/L.
Above 7: basic. pOH = 14 − pH = 5.717.
This is an example. Type your own values to start a new problem.
Kₐ
Parent acid’s Kₐ
example
Kb
Anion’s base constant
calculated
C
Salt concentration
example · M
[OH⁻]
Hydroxide ion concentration
calculated · M
pOH
pOH
calculated
pH
pH
calculated
Formulas
Kb
=
1 ×
10
-14
Kₐ
1.47 ×
10
-11
=
1 ×
10
-14
6.8 ×
10
-4
Kb
=
[OH⁻]²
÷
(
C
−
[OH⁻])
1.47 ×
10
-11
=
(1.92 ×
10
-6
)
2
÷
(0.25
−
(1.92
×
10
-6
))
pOH = −log₁₀([OH⁻])
5.72
=
−log₁₀(1.92
×
10
-6
)
pH = 14.00 − pOH
8.28 = 14.00 − 5.72
Assumptions
•
The salt dissolves fully. Its anion A⁻ is the conjugate base of a weak acid, so A⁻ + H₂O ⇌ HA + OH⁻ makes the water basic.
•
The anion’s Kb is Kw ÷ Kₐ of its parent acid, with Kw = 1 × 10⁻¹⁴ at 25 °C. Ions like Na⁺ and K⁺ leave the pH alone.
•
Kb = y² ÷ (C − y) is solved exactly by the quadratic formula. The shortcut y ≈ √(KbC) is fine when it lands within 5%.
•
Water’s own OH⁻ is left out, which is fine while y is above 10⁻⁶ M.
Step-by-step
WE KNOW
Kₐ = 6.8 × 10⁻⁴, C = 0.25 M
FIND
anion’s base constant (Kb), hydroxide ion concentration ([OH⁻]), pOH (pOH), pH (pH)
Step 1 · Find anion’s base constant (Kb)
Use Kb = (1 × 10⁻¹⁴) ÷ Kₐ
For an acid and its conjugate base, Kₐ × Kb = Kw. The weaker the acid, the stronger its anion is as a base.
Kb
=
10
-14
6.8 ×
10
-4
Kb
=
1.47 ×
10
-11
Step 2 · Find hydroxide ion concentration ([OH⁻])
Use Kb = [OH⁻]² ÷ (C − [OH⁻])
Kb
=
y
2
C
− y
is
y
2
+
Kb·y
−
Kb·
C
=
0.
Its
positive
root
is
written
as
2K
bC
Kb +
√
Kb² + 4KbC
so
a
tiny
y
keeps
its
digits.
[OH⁻]
=
2
×
Kb
×
C
Kb +
√
Kb² + 4 × Kb ×
C
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
÷
(1.47
×
10
-11
+
√(
(1.47 ×
10
-11
)
2
+
4
×
(1.47
×
10
-11
)
×
0.25))
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
÷
((1.47
×
10
-11
)
+
√((2.1609
×
10
-22
)
+
4
×
(1.47
×
10
-11
)
×
0.25))
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
÷
((1.47
×
10
-11
)
+
√((2.1609
×
10
-22
)
+
(5.88
×
10
-11
)
×
0.25))
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
÷
((1.47
×
10
-11
)
+
√((2.1609
×
10
-22
)
+
(1.47
×
10
-11
)))
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
(1.47 ×
10
-11
) +
√
1.47 ×
10
-11
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
(1.47 ×
10
-11
) + (3.834 ×
10
-6
)
[OH⁻]
=
2
×
(1.47
×
10
-11
)
×
0.25
3.834 ×
10
-6
[OH⁻]
=
(2.94
×
10
-11
)
×
0.25
3.834 ×
10
-6
[OH⁻]
=
7.35 ×
10
-12
3.834 ×
10
-6
[OH⁻]
=
1.92 ×
10
-6
M
→
The
shortcut
√
KbC
gives
1.92 ×
10
-6
M,
under
0.1%
off:
close
enough
to
use
Step 3 · Find pOH (pOH)
Use pOH = −log₁₀([OH⁻])
pOH counts the powers of ten in [OH⁻].
pOH
=
−log₁₀(1.92
×
10
-6
)
pOH = 5.72
Step 4 · Find pH (pH)
Use pH = 14.00 − pOH
In
water
at
25
°
C
,
pH
and
pOH
add
to
14.00.
pH = 14.00 − 5.72
pH = 8.28
Check
Put the numbers back in. Both sides should match.
1.47 ×
10
-11
=
10
-14
6.8 ×
10
-4
✓
1.47 ×
10
-11
=
(1.92 ×
10
-6
)
2
÷
(0.25
−
(1.92
×
10
-6
))
✓
5.72
=
−log₁₀(1.92
×
10
-6
)
✓
8.28 = 14.00 − 5.72 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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pH at the equivalence point
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