Answer

Complete and balance the following Brønsted-Lowry acid-base reactions and find the value for K_c for the reaction.

0) NH₃(aq) + HCN(aq)

NH₃(aq) + HCN(aq) → ← NH₄⁺(aq) + CN^–(aq)

K_c = K_a(HCN)/K_a(NH₄⁺) = (6.2×10^–10)/(5.6×10^–10) = 1.1

1) NH₃(aq) + HOCN(aq)

NH₃(aq) + HOCN(aq) → NH₄⁺(aq) + OCN^–(aq)

K_c = K_a(HOCN)/K_a(NH₄⁺) = (3.5×10^–4)/(5.6×10^–10) = 6.3×10⁵

2) NH₃(aq) + HF(aq)

NH₃(aq) + HF(aq) → NH₄⁺(aq) + F^–(aq)

K_c = K_a(HF)/K_a(NH₄⁺) = (6.6×10^–4)/(5.6×10^–10) = 1.2×10⁶

3) NH₃(aq) + HOCl(aq)

NH₃(aq) + HOCl(aq) → ← NH₄⁺(aq) + OCl^–(aq)

K_c = K_a(HOCl)/K_a(NH₄⁺) = (2.9×10^–8)/(5.6×10^–10) = 52

4) NH₃(aq) + HNO₂(aq)

NH₃(aq) + HNO₂(aq) → NH₄⁺(aq) + NO₂^–(aq)

K_c = K_a(HNO₂)/K_a(NH₄⁺) = (7.2×10^–4)/(5.6×10^–10) = 1.3×10⁶

5) N₂H₄(aq) + CH₃CO₂H(aq)

N₂H₄(aq) + CH₃CO₂H(aq) → N₂H₅⁺(aq) + CH₃CO₂^–(aq)

K_c = K_a(CH₃CO₂H)/K_a(N₂H₅⁺) = (1.8×10^–5)/(1.2×10^–8) = 1.5×10³

6) N₂H₄(aq) + HClO₂(aq)

N₂H₄(aq) + HClO₂(aq) → N₂H₅⁺(aq) + ClO₂^–(aq)

K_c = K_a(HClO₂)/K_a(N₂H₅⁺) = (1.1×10^–2)/(1.2×10^–8) = 9.2×10⁵

7) N₂H₄(aq) + HOI(aq)

N₂H₄(aq) + HOI(aq) → ← N₂H₅⁺(aq) + OI^–(aq)

K_c = K_a(HOI)/K_a(N₂H₅⁺) = (2.3×10^–11)/(1.2×10^–8) = 1.9×10^–3

8) N₂H₄(aq) + HIO₃(aq)

N₂H₄(aq) + HIO₃(aq) → N₂H₅⁺(aq) + IO₃^–(aq)

K_c = K_a(HIO₃)/K_a(N₂H₅⁺) = (1.6×10^–1)/(1.2×10^–8) = 1.3×10⁷

9) N₂H₄(aq) + HN₃(aq)

N₂H₄(aq) + HN₃(aq) → N₂H₅⁺(aq) + N₃^–(aq)

K_c = K_a(HN₃)/K_a(N₂H₅⁺) = (1.9×10^–5)/(1.2×10^–8) = 1.6×10³