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Weak Acids and Ka: Quiz

Seven questions on Ka, pKa and weak acid and base calculations, with explanations.

7 questionsAdvanced

7 questions

  1. Question 1EasyWhich acid is the strongest?
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    Answer: HF, Ka=6.8×10−4K_\text{a} = 6.8 \times 10^{-4}

    The largest Ka means the greatest extent of ionization, so HF is the strongest of these weak acids.

  2. Question 2EasyAn acid has Ka=1.8×10−5K_\text{a} = 1.8 \times 10^{-5}. What is its pKa?
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    Answer: 4.74

    pKa = −log(1.8 × 10⁻⁵) = 4.74.

  3. Question 3MediumWhat is the pH of 0.20 M acetic acid (Ka=1.8×10−5K_\text{a} = 1.8 \times 10^{-5})?
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    Answer: 2.72

    [HX3OX+]≈1.8×10−5×0.20=1.9×10−3[\ce{H3O+}] \approx \sqrt{1.8 \times 10^{-5} \times 0.20} = 1.9 \times 10^{-3} M, so pH = 2.72. The answer 0.70 treats acetic acid as a strong acid.

  4. Question 4HardA 0.10 M weak acid has pH 3.00. What is its KaK_\text{a}?
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    Answer: 1.0×10−51.0 \times 10^{-5}

    [HX3OX+]=[AX−]=1.0×10−3[\ce{H3O+}] = [\ce{A-}] = 1.0 \times 10^{-3} M; Ka=(1.0×10−3)2÷0.099=1.0×10−5K_\text{a} = (1.0 \times 10^{-3})^2 \div 0.099 = 1.0 \times 10^{-5}.

  5. Question 5MediumWhat is the pH of 0.20 M ammonia (Kb=1.8×10−5K_\text{b} = 1.8 \times 10^{-5})?
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    Answer: 11.28

    [OHX−]≈1.9×10−3[\ce{OH-}] \approx 1.9 \times 10^{-3} M, pOH = 2.72, pH = 14.00 − 2.72 = 11.28. The answer 2.72 reports the pOH.

  6. Question 6MediumFor a conjugate acid–base pair at 25 °C, Ka×KbK_\text{a} \times K_\text{b} equals…
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    Answer: 1.0×10−141.0 \times 10^{-14}

    Ka×Kb=Kw=1.0×10−14K_\text{a} \times K_\text{b} = K_\text{w} = 1.0 \times 10^{-14} at 25 °C.

  7. Question 7EasyWhy can C−xC - x often be replaced by CC in weak acid calculations?
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    Answer: Because only a small fraction of a weak acid ionizes

    For a weak acid, x is usually tiny compared with C (under 5%), so C − x ≈ C. Always check this.