SME
General Chemistry II · Problem type
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.”
Diagram
pH
=
pKₐ
+
log(
n(A⁻)
n(HA)
)
pH
=
3.17
+
log(
0.25
0.2
)
=
3.27
The shaded band is pKₐ ± 1 (a ratio from 1 : 10 to 10 : 1), where it buffers well.
Adding 0.02 mol of strong base turns that much HA into A⁻: pH 3.35.
This is an example. Type your own values to start a new problem.
pH
=
pKₐ
+
log(
n(A⁻) (mol)
÷
n(HA) (mol)
)
a
Strong acid added (− for base)
example · mol
pH′
pH after
calculated
Formulas
pH
=
pKₐ
+
log₁₀(
n(A⁻)
n(HA)
)
3.27
=
3.17
+
log₁₀(
0.25
0.2
)
pH′
=
pKₐ
+
log₁₀(
n(A⁻) −
a
n(HA) +
a
)
3.35
=
3.17
+
log₁₀(
0.25 − (−0.02)
0.2 + (−0.02)
)
Assumptions
•
A buffer holds a weak acid HA and its conjugate base A⁻. Henderson–Hasselbalch gives its pH from pKₐ and their ratio.
•
Both sit in the same volume, so the ratio of moles equals the ratio of concentrations.
•
Strong acid turns that much A⁻ into HA. Strong base is typed as a negative amount and turns HA into A⁻.
•
Both amounts stay much larger than what is added, and neither is over 100 times the other, so the pH stays within 2 of pKₐ.
Step-by-step
WE KNOW
pKₐ = 3.17, n(HA) = 0.2 mol, n(A⁻) = 0.25 mol, a = −0.02 mol
FIND
pH before (pH), pH after (pH′)
Step 1 · Find pH before (pH)
Use pH = pKₐ + log₁₀(n(A⁻) ÷ n(HA))
Henderson–Hasselbalch: add the log of the base-to-acid ratio to pKₐ.
pH
=
3.17
+
log₁₀(
0.25
0.2
)
pH = 3.17 + log₁₀(1.25)
pH = 3.17 + 0.09691
pH = 3.27
Step 2 · Find pH after (pH′)
Use pH′ = pKₐ + log₁₀((n(A⁻) − a) ÷ (n(HA) + a))
Strong acid turns that much A⁻ into HA, and strong base does the reverse. Change both amounts, then use the new ratio.
pH′
=
3.17
+
log₁₀(
0.25 − (−0.02)
0.2 + (−0.02)
)
pH′
=
3.17
+
log₁₀(
0.27
0.18
)
pH′ = 3.17 + 0.1761
pH′ = 3.35
Check
Put the numbers back in. Both sides should match.
3.27
=
3.17
+
log₁₀(
0.25
0.2
)
✓
3.35
=
3.17
+
log₁₀(
0.25 − (−0.02)
0.2 + (−0.02)
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Acid–base equilibria and buffers
Main lesson
pH of a weak base
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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.”