Unit 8 Practice: Medium
Acids and Bases · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
Calculate the pH of 0.0100 M HCl.
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Answer: 2.00
Explanation: HCl is a strong acid, so:
[H₃O⁺] ≈ 0.0100 M
Then:
pH = −log(0.0100)
= 2.00
Question 2
Calculate the pH of 0.00100 M NaOH at 25°C.
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Answer: 11.00
Work:
NaOH is a strong base:
[OH⁻] = 1.00 × 10⁻³ M
So:
pOH = 3.00
Then:
pH = 14.00 − 3.00 = 11.00
Question 3
Calculate [H₃O⁺] if pH = 4.50.
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Answer: 3.16 × 10⁻⁵ M
Work:
[H₃O⁺] = 10⁻pH
= 10⁻⁴·⁵⁰
= 3.16 × 10⁻⁵ M
Question 4
Calculate [OH⁻] if pOH = 2.30.
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Answer: 5.01 × 10⁻³ M
Work:
[OH⁻] = 10⁻²·³⁰
= 5.01 × 10⁻³ M
Question 5
A 0.0200 M solution of Ca(OH)₂ dissociates completely. Find [OH⁻].
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Answer: 0.0400 M
Explanation: Each Ca(OH)₂ produces 2 OH⁻:
Ca(OH)₂ → Ca²⁺ + 2OH⁻
So:
[OH⁻] = 2(0.0200) = 0.0400 M
Question 6
Write the acid-ionization equation for acetic acid, CH₃COOH.
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Answer:
CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻
Explanation: Acetic acid donates H⁺ to water.
Question 7
Write Ka for acetic acid.
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Answer:
Ka = [H₃O⁺][CH₃COO⁻] / [CH₃COOH]
Explanation: Liquid water is omitted from the equilibrium expression.
Question 8
Write Kb for NH₃.
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Answer:
Kb = [NH₄⁺][OH⁻] / [NH₃]
Explanation:
Reaction:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Question 9
A weak acid HA has:
Ka = 1.0 × 10⁻⁵
Is HA stronger or weaker than an acid with:
Ka = 1.0 × 10⁻³?
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Answer: HA is weaker.
Explanation: The second acid has the larger Ka and therefore ionizes more strongly.
Question 10
If Ka for HA is 1.0 × 10⁻⁵, find Kb for A⁻ at 25°C.
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Answer: 1.0 × 10⁻⁹
Work:
KaKb = Kw
So:
Kb = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻⁵
= 1.0 × 10⁻⁹
Question 11
If an acid becomes stronger, what happens to its conjugate base?
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Answer: The conjugate base becomes weaker.
Explanation: Strong acids lose H⁺ easily.
Their conjugate bases therefore have little tendency to take H⁺ back.
Question 12
What is percent ionization?
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Answer:
Percent ionization = (amount ionized / initial acid concentration) × 100%
Explanation: For a weak monoprotic acid:
% ionization = ([H₃O⁺]eq / [HA]initial) × 100%
Question 13
What generally happens to percent ionization of a weak acid when the solution is diluted?
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Answer: Percent ionization increases.
Explanation: Dilution favors additional ionization of the weak acid.
The fraction of acid molecules that ionize increases, even though the actual [H₃O⁺] may decrease.
Question 14
Is NaCl(aq) usually acidic, basic, or approximately neutral?
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Answer: Approximately neutral
Explanation: Na⁺ comes from a strong base and Cl⁻ comes from a strong acid.
Neither ion reacts significantly with water.
Question 15
Is NH₄Cl(aq) acidic, basic, or neutral?
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Answer: Acidic
Explanation: NH₄⁺ is the conjugate acid of weak base NH₃.
It can donate H⁺ to water.
Cl⁻ is essentially neutral.
Question 16
Is NaCH₃COO(aq) acidic, basic, or neutral?
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Answer: Basic
Explanation: CH₃COO⁻ is the conjugate base of weak acid CH₃COOH.
It reacts with water to produce OH⁻.
Question 17
What is a buffer?
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Answer: A solution that resists large pH changes when small amounts of acid or base are added.
Explanation: A buffer usually contains:
a weak acid and its conjugate base
or
a weak base and its conjugate acid
Question 18
Write the Henderson-Hasselbalch equation.
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Answer:
pH = pKa + log([A⁻]/[HA])
Explanation: It is useful for buffer systems made from a weak acid and its conjugate base.
Question 19
In a buffer, what happens when [A⁻] = [HA]?
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Answer: pH = pKa
Explanation: Because:
log(1) = 0
So:
pH = pKa
Question 20
At the half-equivalence point of a weak acid-strong base titration, what relationship is true?
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Answer: pH = pKa
Explanation: At half-equivalence:
[HA] = [A⁻]
so the Henderson-Hasselbalch equation gives pH = pKa.