Unit 8 Practice: Hard
Acids and Bases · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
A 0.100 M weak acid HA has:
Ka = 1.0 × 10⁻⁵
Estimate [H₃O⁺].
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Answer: 1.0 × 10⁻³ M
Work:
Reaction:
HA ⇌ H₃O⁺ + A⁻
Let x = [H₃O⁺].
Then:
Ka = x²/(0.100 − x)
If x is small:
1.0 × 10⁻⁵ ≈ x²/0.100
x² = 1.0 × 10⁻⁶
x = 1.0 × 10⁻³ M
Question 2
Find the pH of the solution in Question 1.
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Answer: 3.00
Work:
pH = −log(1.0 × 10⁻³)
= 3.00
Question 3
Check whether the small-x approximation in Question 1 is reasonable.
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Answer: Yes.
Work:
Percent change:
(0.0010 / 0.100)(100)
= 1.0%
Explanation: Because the change is much less than about 5%, the approximation is reasonable.
Question 4
A 0.200 M weak acid has pH = 2.70. Estimate its Ka.
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Answer: About 2.0 × 10⁻⁵
Work:
First:
[H₃O⁺] = 10⁻²·⁷⁰
≈ 2.0 × 10⁻³ M
So:
x = 2.0 × 10⁻³
Then:
Ka = x²/(0.200 − x)
≈ (4.0 × 10⁻⁶)/(0.198)
≈ 2.0 × 10⁻⁵
Question 5
A 0.100 M NH₃ solution has:
Kb = 1.8 × 10⁻⁵
Estimate [OH⁻].
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Answer: 1.34 × 10⁻³ M
Work:
For:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Assume x is small:
Kb ≈ x²/0.100
1.8 × 10⁻⁵ = x²/0.100
x² = 1.8 × 10⁻⁶
x = 1.34 × 10⁻³ M
Question 6
Find the pH of the NH₃ solution in Question 5.
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Answer: About 11.13
Work:
pOH = −log(1.34 × 10⁻³)
≈ 2.87
Then:
pH = 14.00 − 2.87
= 11.13
Question 7
Why is HF a weaker acid than HCl even though F is more electronegative than Cl?
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Answer: The H—F bond is much stronger than the H—Cl bond.
Explanation: For binary acids in the same group, bond strength is very important.
HF holds onto H⁺ strongly because the H—F bond is short and strong.
HCl ionizes much more completely.
Question 8
For oxyacids with the same central atom, why does adding more oxygen atoms usually increase acid strength?
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Answer: The additional oxygen atoms help stabilize the conjugate base.
Explanation: More electronegative oxygen atoms pull electron density away and often allow the negative charge of the conjugate base to be better distributed.
A more stable conjugate base corresponds to a stronger acid.
Question 9
Which is generally the stronger acid: HClO or HClO₄?
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Answer: HClO₄
Explanation: HClO₄ has more oxygen atoms.
Its conjugate base is much more stabilized by electron withdrawal and charge delocalization.
Question 10
A buffer contains 0.30 mol HA and 0.30 mol A⁻. If pKa = 4.76, find the pH.
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Answer: 4.76
Work:
pH = pKa + log(A⁻/HA)
= 4.76 + log(0.30/0.30)
= 4.76 + log(1)
= 4.76
Question 11
A buffer contains 0.20 mol HA and 0.40 mol A⁻. If pKa = 4.76, find the pH.
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Answer: About 5.06
Work:
pH = 4.76 + log(0.40/0.20)
= 4.76 + log(2)
≈ 4.76 + 0.301
≈ 5.06
Question 12
A buffer initially contains 0.30 mol HA and 0.30 mol A⁻. Then 0.050 mol HCl is added. What happens before using Henderson-Hasselbalch?
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Answer: H⁺ reacts with A⁻ first.
Explanation:
Reaction:
H⁺ + A⁻ → HA
New amounts:
A⁻:
0.30 − 0.050 = 0.250 mol
HA:
0.30 + 0.050 = 0.350 mol
Only after this stoichiometric reaction should the buffer equation be used.
Question 13
Using Question 12 and pKa = 4.76, estimate the new pH.
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Answer: About 4.61
Work:
pH = 4.76 + log(0.250/0.350)
Ratio:
0.250/0.350 ≈ 0.714
log(0.714) ≈ −0.146
Therefore:
pH ≈ 4.76 − 0.146
≈ 4.61
Question 14
Why does a buffer resist pH change?
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Answer: Its weak acid can consume added OH⁻, while its conjugate base can consume added H⁺.
Explanation: The buffer converts added strong acid or base into weaker species.
That keeps [H₃O⁺] from changing dramatically.
Question 15
A weak acid is titrated with a strong base. Is the pH at equivalence less than, equal to, or greater than 7 at 25°C?
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Answer: Greater than 7
Explanation: At equivalence, the weak acid has been converted mostly to its conjugate base.
That conjugate base reacts with water:
A⁻ + H₂O ⇌ HA + OH⁻
producing a basic solution.
Question 16
A weak base is titrated with a strong acid. Is the pH at equivalence less than, equal to, or greater than 7 at 25°C?
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Answer: Less than 7
Explanation: At equivalence, the weak base has been converted to its conjugate acid.
That species produces H₃O⁺ in water.
Question 17
During a weak acid-strong base titration, what species dominate before equivalence but after some base has been added?
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Answer: Weak acid HA and conjugate base A⁻
Explanation: Some HA has been neutralized into A⁻, but excess HA still remains.
The solution is therefore a buffer.
Question 18
A 25.0 mL sample of 0.100 M weak acid HA is titrated with 0.100 M NaOH. At what volume of NaOH is the half-equivalence point reached?
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Answer: 12.5 mL
Work:
Initial acid moles:
0.0250 L × 0.100 M = 0.00250 mol
Equivalence requires the same moles NaOH:
0.00250 mol
At 0.100 M:
V = 0.00250 / 0.100 = 0.0250 L
= 25.0 mL equivalence volume
Half-equivalence:
12.5 mL
Question 19
For the titration in Question 18, if Ka = 1.8 × 10⁻⁵, what is the pH at half-equivalence?
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Answer: About 4.74
Work:
At half-equivalence:
pH = pKa
So:
pKa = −log(1.8 × 10⁻⁵)
≈ 4.74
Question 20
Why should an indicator’s transition range overlap the steep portion of a titration curve?
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Answer: So the visible endpoint occurs close to the true equivalence point.
Explanation: Indicators change color over a limited pH range.
If that range lies inside the sharp pH change near equivalence, only a very small titrant-volume difference separates the endpoint from the equivalence point.