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Ka and Kb

Chemistry • Acid–Base Equilibria • Knowledge-Based • 5 questions

Ka and Kb Knowledge-Based Download PDF

Chemistry · Acids & Bases · Chemistry · Acid–Base Equilibria

5 Chemistry practice questions on Ka and Kb. Questions that assess core concepts and understanding.

Questions

  1. 1.

    What is the pH of a 0.10 M aqueous solution of a weak base, B, with a base dissociation constant of Kb=1.0×10−5K_b = 1.0 \times 10^{-5} at 25 °C?

    • A) 3.00
    • B) 8.00
    • C) 9.00
    • D) 11.00
  2. 2.

    A 0.10 M aqueous solution of a weak monoprotic acid, HA, has a pH of 4.00 at 25 °C. What is the acid dissociation constant, KaK_a, of this acid?

    • A) 1.0×10−71.0 \times 10^{-7}
    • B) 1.0×10−41.0 \times 10^{-4}
    • C) 1.0×10−81.0 \times 10^{-8}
    • D) 1.0×10−31.0 \times 10^{-3}
  3. 3.

    What is the equilibrium constant, KK, at 25 °C for the reaction HF(aq)+NH3(aq)⇌F−(aq)+NH4+(aq)HF(aq) + NH_3(aq) \rightleftharpoons F^-(aq) + NH_4^+(aq), given that Ka(HF)=6.8×10−4K_a(HF) = 6.8 \times 10^{-4} and Kb(NH3)=1.8×10−5K_b(NH_3) = 1.8 \times 10^{-5}?

    • A) 3.8×1013.8 \times 10^1
    • B) 1.2×10−81.2 \times 10^{-8}
    • C) 8.2×10−78.2 \times 10^{-7}
    • D) 1.2×1061.2 \times 10^6
  4. 4.

    The acid dissociation constants for four weak monoprotic acids at 25 °C are given as HXHX (Ka=7.2×10−4K_a = 7.2 \times 10^{-4}), HYHY (Ka=3.5×10−11K_a = 3.5 \times 10^{-11}), HZHZ (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}), and HQHQ (Ka=4.9×10−8K_a = 4.9 \times 10^{-8}). Which of the conjugate bases has the largest base dissociation constant (KbK_b) at this temperature?

    • A) X−X^-
    • B) Y−Y^-
    • C) Z−Z^-
    • D) Q−Q^-
  5. 5.

    Hypochlorous acid (HClOHClO) has an acid dissociation constant of Ka=3.0×10−8K_a = 3.0 \times 10^{-8} at 25 °C. What is the value of the base dissociation constant (KbK_b) for its conjugate base, the hypochlorite ion (ClO−ClO^-), at the same temperature?

    • A) 3.0×10−63.0 \times 10^{-6}
    • B) 3.0×10−223.0 \times 10^{-22}
    • C) 3.3×10−73.3 \times 10^{-7}
    • D) 3.3×10−83.3 \times 10^{-8}