What is the approximate value of the K for the neutralization of nitrous acid with ammonia? Ka for the acid is 0.00045 and Kb for the base is 0.000018.
Do I multiply Ka*Kb?
2 answers
I think that is the hydrolysis constant which is Kw/(KaKb).
Frist write out what is given:
HNO2 ⇌ H+ + NO2-
Ka = 0.00045 = [H+]*[NO2-]/[HNO2]
and:
NH3 + H2O ⇌ NH4+ + OH-
Kb = 0.000018 = [NH4+]*[OH-]/[NH3]
What is asked for is the K value for:
HNO2 + NH3 ⇌ NH4+ + NO2-
K = [NH4+]*[NO2-]/{ [HNO2]*[NH3] }
= { [NH4+]*[OH-]/[NH3] }*{ [H+]*[NO2-]/[HNO2] }/{ [H+]*[OH-] }
= Kb*Ka/Kw
where Kw = [H+]*[OH-] = 10^-14.
Is that very easy? Please finally do the very simple math.
HNO2 ⇌ H+ + NO2-
Ka = 0.00045 = [H+]*[NO2-]/[HNO2]
and:
NH3 + H2O ⇌ NH4+ + OH-
Kb = 0.000018 = [NH4+]*[OH-]/[NH3]
What is asked for is the K value for:
HNO2 + NH3 ⇌ NH4+ + NO2-
K = [NH4+]*[NO2-]/{ [HNO2]*[NH3] }
= { [NH4+]*[OH-]/[NH3] }*{ [H+]*[NO2-]/[HNO2] }/{ [H+]*[OH-] }
= Kb*Ka/Kw
where Kw = [H+]*[OH-] = 10^-14.
Is that very easy? Please finally do the very simple math.
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