AP Chemistry · Topic 8.8

Properties of Buffers Practice

Part of Acids and Bases.(SAP-10.B)

Practice questions

5

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Sample questions

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  1. Sample 1difficulty 2/5

    A buffer typically consists of

    • A

      Weak acid + its conjugate base (or weak base + conjugate acid)

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    • B

      Two strong acids

    • C

      Pure water

    • D

      Strong acid + strong base

    Why

    Both partners present in comparable amounts to neutralize added acid or base.

  2. Sample 2difficulty 3/5

    CH3COOH/CH3COO- buffer pKa = 4.74 pH = pKa + log([A-]/[HA]) [A-]/[HA] = 1 -> pH = ?

    What is the pH when the conjugate base/acid ratio = 1?

    • A

      7.00

    • B

      4.74

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    • C

      9.26

    • D

      5.74

    Why

    log(1) = 0, so pH = pKa = 4.74.

  3. Sample 3difficulty 3/5

    For a pH-5 buffer, choose a weak acid with pKa

    • A

      ≈ 14

    • B

      ≈ 9

    • C

      ≈ 2

    • D

      ≈ 5

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    Why

    Best buffer near pH = pKa, ideally within ±1 unit.

  4. Sample 4difficulty 4/5

    A buffer is made from 0.40 M NH3NH_3 and 0.30 M NH4ClNH_4Cl. What is the pH? (Kb(NH3)=1.8×105K_b(NH_3) = 1.8\times 10^{-5})

    • A

      pH9.38pH \approx 9.38

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    • B

      pH9.26pH \approx 9.26

    • C

      pH4.62pH \approx 4.62

    • D

      pH8.92pH \approx 8.92

    Why

    pKa(NH4+)pK_a(NH_4^+) = 14 - pKb(NH3)pK_b(NH_3) = 14 - 4.74 = 9.26. pH = pKa+log([NH3]/[NH4+])pK_a + \log([NH_3]/[NH_4^+]) = 9.26 + log(0.40/0.30)\log(0.40/0.30) = 9.26 + 0.125 = 9.38.

  5. Sample 5difficulty 4/5

    A buffer contains 0.20 M HF and 0.30 M NaF. The pH is (KaK_a(HF) = 7.2×1047.2\times10^{-4}, log7.2=0.86\log 7.2 = 0.86)

    • A

      3.14

    • B

      3.44

    • C

      3.32 (pKa+log([F]/[HF])pK_a + \log([F^-]/[HF]))

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    • D

      10.86

    Why

    pH = pKa+log([F]/[HF])pK_a + \log([F^-]/[HF]) = log(7.2×104)+log(0.30/0.20)-\log(7.2\times10^{-4}) + \log(0.30/0.20) = 3.143 + 0.176 = 3.32.