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CHM 112
Section 1

Homework Assignment #20
Due March 29, 2006 by 8:05 a.m.

Practice (not to be turned in): Practice reactions 1-20

Complete and balance the following Brønsted-Lowry acid-base reactions.

0) NH3(aq) + HCN(aq)

1) NH3(aq) + HOCN(aq)

2) NH3(aq) + HF(aq)

3) NH3(aq) + HOCl(aq)

4) NH3(aq) + HNO2(aq)

5) N2H4(aq) + CH3CO2H(aq)

6) N2H4(aq) + HClO2(aq)

7) N2H4(aq) + HOI(aq)

8) N2H4(aq) + HIO3(aq)

9) N2H4(aq) + HN3(aq)



Answers:

In each case, the acid is written second and acts as the hydrogen ion donor. To establish if the reaction is an equilibrium or goes essentially to completion, the equilibrium constant must be found from Kc = Ka(acid)/Ka(conjugate acid).



0) NH3(aq) + HCN(aq) NH4+(aq) + CN(aq)

Ka(HCN) = 6.2×10–10

Ka(NH4)+ = 5.6×10–10

Kc = (6.2×10–10)/(5.6×10–10) = 1.1 < 100, equilibrium.



1) NH3(aq) + HOCN(aq) NH4+(aq) + OCN(aq)

Ka(HOCN) = 3.5×10–4

Ka(NH4+) = 5.6×10–10

Kc = (3.5×10–4)/(5.6×10–10) = 6.3×105 > 100, completion.



2) NH3(aq) + HF(aq) NH4+(aq) + F(aq)

Ka(HF) = 6.6×10–4

Ka(NH4+) = 5.6×10–10

Kc = (6.6×10–4)/(5.6×10–10) = 1.2×106 > 100, completion.



3) NH3(aq) + HOCl(aq) NH4+(aq) + OCl(aq)

Ka(HOCl) = 2.9×10–8

Ka(NH4+) = 5.6×10–10

Kc = (2.9×10–8)/(5.6×10–10) = 52 < 100, equilibrium.



4) NH3(aq) + HNO2(aq) NH4+(aq) + NO2(aq)

Ka(HNO2) = 7.2×10–4

Ka(NH4+) = 5.6×10–10

Kc = (7.2×10–4)/(5.6×10–10) = 1.3×106 > 100, completion.



5) N2H4(aq) + CH3CO2H(aq) N2H5+(aq) + CH3CO2(aq)

Ka(CH3CO2H) = 1.8×10–5

Ka(N2H5+) = 1.2×10–8

Kc = (1.8×10–5)/(1.2×10–8) = 1500 > 100, completion.



6) N2H4(aq) + HClO2(aq) N2H5+(aq) + ClO2(aq)

Ka(HClO2) = 1.1×10–2

Ka(N2H5+) = 1.2×10–8

Kc = (1.1×10–2)/(1.2×10–8) = 9.2×105 > 100, completion.



7) N2H4(aq) + HOI(aq) N2H5+(aq) + OI(aq)

Ka(HOI) = 2.3×10–11

Ka(N2H5+) = 1.2×10–8

Kc = (2.3×10–11)/(1.2×10–8) = 1.9×10–3 < 100, equilibrium.



8) N2H4(aq) + HIO3(aq) N2H5+(aq) + IO3(aq)

Ka(HIO3) = 1.6×10–1

Ka(N2H5+) = 1.2×10–8

Kc = (1.6×10–1)/(1.2×10–8) = 1.3×107 > 100, completion.



9) N2H4(aq) + HN3(aq) N2H5+(aq) + N3(aq)

Ka(HN3) = 1.9×10–5

Ka(N2H5+) = 1.2×10–8

Kc = (1.9×10–5)/(1.2×10–8) = 1600 > 100, completion.