Acids and Bases

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What you have to know

14 practice questions

1. Calculate the pH of a 0.025 M solution of HCl (a strong acid).
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   Answer. pH = 1.60

   HCl dissociates completely, so [H⁺] = 0.025 M. pH = −log(0.025) = 1.60.
2. What is the pH of a 0.010 M NaOH solution?
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   Answer. pH = 12.00

   NaOH is a strong base and fully dissociates, giving [OH⁻] = 0.010 M, so pOH = −log(0.010) = 2.00. Then pH = 14.00 − 2.00 = 12.00.
3. A 0.20 M solution of weak acid HA has Ka = 1.0×10⁻⁵. Calculate its pH.
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   Answer. pH ≈ 2.85

   Set up an ICE table: x²/(0.20 − x) ≈ x²/0.20 = 1.0×10⁻⁵ (approximation valid since Ka is small). Solving gives x = [H⁺] = 1.4×10⁻³ M, so pH = −log(1.4×10⁻³) = 2.85.
4. 0.10 M solutions of HCl and HF are prepared separately. Which solution has the higher pH, and why?
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   Answer. The HF solution has the higher pH.

   HCl is a strong acid and dissociates completely, giving [H⁺] = 0.10 M. HF is a weak acid and only partially ionizes, so its [H⁺] is much lower than 0.10 M, resulting in a higher pH for the HF solution.
5. Which of the following oxoacids is the strongest acid?
   • HClO
   • HClO2
   • HClO3
   • HClO4
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   Answer. HClO4

   Acid strength increases with the number of oxygen atoms bonded to the central atom, since more oxygens withdraw electron density and stabilize the negative charge on the conjugate base. HClO4 has the most oxygens of the four choices.
6. Explain why HI is a stronger acid than HF, even though fluorine is more electronegative than iodine.
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   Answer. HI is stronger because the H–I bond is much weaker than the H–F bond.

   Bond strength decreases going down a group, making the H–I bond easier to break and release H⁺. For binary acids, this bond-strength effect dominates over the electronegativity trend, so HI ionizes more readily than HF.
7. A buffer is prepared by combining 0.15 M NH₃ and 0.10 M NH₄Cl (Kb of NH₃ = 1.8×10⁻⁵). Calculate the pH of the buffer.
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   Answer. pH ≈ 9.43

   Find Ka of NH₄⁺: Ka = Kw/Kb = (1.0×10⁻¹⁴)/(1.8×10⁻⁵) = 5.6×10⁻¹⁰, so pKa = 9.25. Apply Henderson-Hasselbalch: pH = 9.25 + log(0.15/0.10) = 9.25 + 0.18 = 9.43.
8. In a titration of a weak acid with a strong base, what is true about the pH at the half-equivalence point?
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   Answer. pH equals the pKa of the weak acid.

   At the half-equivalence point, exactly half of the original HA has been converted to A⁻, so [HA] = [A⁻]. Henderson-Hasselbalch then gives pH = pKa + log(1) = pKa.
9. How does adding solid sodium acetate (NaCH₃COO) to a solution of acetic acid affect the solution's pH, and why?
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   Answer. The pH increases (the solution becomes less acidic).

   The added acetate ion is a common ion that shifts the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ to the left by Le Chatelier's principle, decreasing [H⁺] and raising the pH.
10. What is the pH at the equivalence point of a titration between a strong acid and a strong base?
    • Less than 7
    • Equal to 7
    • Greater than 7
    • It cannot be determined without concentrations
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    Answer. Equal to 7

    The salt formed (for example NaCl) comes from a strong acid and a strong base, so neither ion hydrolyzes appreciably in water, leaving the solution neutral at pH 7.
11. At the equivalence point of a titration between a weak base and a strong acid, is the pH greater than, less than, or equal to 7? Explain.
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    Answer. Less than 7.

    The salt formed contains the conjugate acid of the weak base (such as NH₄⁺), which hydrolyzes in water to produce H₃O⁺, making the solution acidic at the equivalence point.
12. A 0.10 M solution of a weak monoprotic acid is found to be 3.0% ionized. Calculate Ka.
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    Answer. Ka ≈ 9.3×10⁻⁵

    [H⁺] = 0.030 × 0.10 M = 3.0×10⁻³ M. Ka = [H⁺]²/([HA]₀ − [H⁺]) = (3.0×10⁻³)²/(0.10 − 3.0×10⁻³) = 9.0×10⁻⁶/0.097 = 9.3×10⁻⁵.
13. For a diprotic acid H₂A with Ka1 = 4.5×10⁻³ and Ka2 = 1.7×10⁻⁸, explain why the second ionization can usually be ignored when calculating the pH of an H₂A solution.
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    Answer. Because Ka2 is many orders of magnitude smaller than Ka1, the second ionization contributes negligible extra H⁺.

    The first ionization already produces most of the H⁺ present, and the H⁺ from step one further suppresses the second ionization through the common ion effect, so [H⁺] is essentially determined by Ka1 alone.
14. 50.0 mL of 0.20 M CH₃COOH is titrated with 0.10 M NaOH. What volume of NaOH is required to reach the equivalence point?
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    Answer. 100.0 mL

    Moles CH₃COOH = 0.0500 L × 0.20 M = 0.0100 mol. At the equivalence point, moles NaOH added must equal moles acid, so volume = 0.0100 mol ÷ 0.10 M = 0.100 L = 100.0 mL.

What people get wrong

1. Treating a weak acid like a strong acid by setting [H⁺] equal to the initial concentration — instead, build an ICE table and solve for x using Ka, since weak acids only partially dissociate.
2. Assuming pH + pOH = 14 always holds — this depends on Kw = 1.0×10⁻¹⁴, which is only true at 25°C; at other temperatures Kw changes and so does the sum.
3. Assuming the half-equivalence point always gives pH = pKa — this shortcut only applies to weak acid/strong base (or weak base/strong acid) titrations, not to strong-strong titrations which have no meaningful 'pKa'.
4. Ranking acid strength purely by electronegativity — for binary acids down a group, decreasing bond strength (not increasing electronegativity) controls the trend, so HI is a stronger acid than HF.
5. Forgetting the common ion effect — adding the conjugate base of a weak acid suppresses its ionization via Le Chatelier's principle, so do not simply reuse the original Ka calculation as if no common ion were present.
6. Assuming every equivalence point occurs at pH 7 — this is true only for strong acid–strong base titrations; weak acid/strong base gives pH above 7, and weak base/strong acid gives pH below 7.