PLEASE SHOW ALL STEPS TO ANSWERING THE PROBLEM AND EXPLAIN WHY FOR CONCEPTUAL QU
ID: 800971 • Letter: P
Question
PLEASE SHOW ALL STEPS TO ANSWERING THE PROBLEM AND EXPLAIN WHY FOR CONCEPTUAL QUESTIONS
1.) Calculate the value of the equilibrium constant for the following reactions:
a) NH3 + H3O+ <--> NH4+ + H2O
b) NO2- + H3O+ <--> HNO2 + H2O
c) NH4+ + OH- <--> NH3 +H2O
d) HNO2 + OH- <--> NO2- + H2O
e) Classify the reaction in a through d.
f) In solving problems, for the reactions a through d, should the final concentration for these reactions be found by treating the reactions as equilibrium or as stoichiometry problems?
ANSWER:
a) 1.8 X 10^9
b) 1.4 X 10^3
c) 5.6 X 10^4
d) 7.2 X 10^10
e) Strong acid - weak base or weak acid- strong base
f) Treated as stoichometry
g) [NH3] = 0.25M, [H+] = negligible, [NH4+]= 0.25 M
2.) Calculate the pH of the following aqueous solutions:
a)1.5*10-5M HCl
b)1.0 mL of sample a in 0.50 L of water.
ANSWER:
a) 4.82
b) 6.94
Explanation / Answer
1)
A) NH3 + H3O+ <--> NH4+ + H2O
the equilibrium constant
K = [ NH4+] [H20] / [ H30+] [NH3]
water is neglected
K= [ NH4+] / [ H30+] [NH3]
NH4+ + H20 ----> NH3 + H30+
Ka = [NH3] [H30+] / [NH4+]
K = 1/ Ka
K = 1/ 5.6 x 10-10
K = 1.7857 x 10^9
b)
NO2- + H3O+ <--> HNO2 + H2O
K = [HN02] / [H30+] [ N02-]
HN02 + H20 -----> N02- + H30+
Ka = [N02-] [H30+] / [HN02]
K = 1/Ka
K = 1/ 7.2 x 10-4
K = 1.8 x 10^3
C)
NH4+ + OH- <--> NH3 +H2O
K = [NH3] / [ Nh4+][OH-]
NH3 + H20 ---> NH4+ + OH
Kb = [ NH4+] [OH-] / [NH3]
K = 1/Kb
K = 1/ 1.8 x 10-5
K = 5.55 x 10^4
D)
HNO2 + OH- <--> NO2- + H2O
K = [N02-] / [HN02] [OH-]
N02- + H20 ----> HN02 + OH-
Kb = [HN02] [OH-] / [N02-]
K = 1/Kb
K = 1/ 1.4 x 10-11
k = 7.2 x 10^ 10
e) NH3 + H3O+ <--> NH4+ + H2O
weak base strong acid
NO2- + H3O+ <--> HNO2 + H2O
strong base strong acid
NH4+ + OH- <--> NH3 +H2O
strong acid and strong base
HNO2 + OH- <--> NO2- + H2O
weak acid strong base
f) they are to be treated as stoiciometry because they are of the form
K = [C] [D] / [A] [B]
here all the volumes get cancelled and only the moles remian .
2) a) pH = -log [H+]
pH = -log 1.5 x 10-5
pH = 4.824
b) moles of Hcl = volume x molarity / 1000
moles of Hcl = 1 x 1.5 x 10-5 / 1000
moles of Hcl = 1.5 x 10-8
Final volume = 501 ml
final conc of Hcl = 1.5 x 10-8 x 1000 / 501
conc of Hcl = 2.994 x 10-8
the conc is very low .So H+ ions in water
So total [H+] = 0.2994 x 10-7 + 10-7
[H+] = 1.2994 x 10-7
pH = -log 1.2994 x 10-7
pH = 6.88
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