(b) Which is the stronger conjugate acid (which behaves as an acid), the ammoniu

Question

(b) Which is the stronger conjugate acid (which behaves as an acid), the ammonium ion or the methylammonium ion?

(c) Given Kb above, use KW to calculate Ka for THEIR conjugate acids (you do this for salt problems) NH4+ and CH3NH3+.

Given that Kb for ammonia, NH3 is 1.8 times 10-5 and that for methylamine,CH3NH2 is 4.4 times 10-4, which is the stronger base? methylamine,CH3NH2ammonia, NH3  Which is the stronger conjugate acid (which behaves as an acid), the ammonium ion or the methylammonium ion? methylammonium ionammonium ion Given Kb above, use KW to calculate Ka for THEIR conjugate acids (you do this for salt problems) NH4+ and CH3NH3+.

Explanation / Answer

(a) The answer is: methylamine, CH3NH2

For weak base: B + H2O <=> BH+ + OH-

Kb = [BH+][OH-]/[B]


Larger Kb => higher OH- concentration => stronger base

Since methylamine has a larger Kb => methylamine, CH3NH2 is the stronger base


(b) The answer is: ammonium ion

Weaker base => stronger conjugate acid

Since ammonia is the weaker base => its conjugate acid ammonium ion is the stronger acid


(c) Ka for NH4+ = Kw/Kb for NH3

= 10^(-14)/1.8 x 10^(-5) = 5.56 x 10^(-10) = 5.6 x 10^(-10)


Ka for CH3NH3+ = Kw/Kb for CH3NH2

= 10^(-14)/4.4 x 10^(-4) = 2.27 x 10^(-11) = 2.3 x 10^(-11)


(d) Kb for CN- = Kw/Ka for HCN

= 10^(-14)/4.9 x 10^(-10) = 2.04 x 10^(-5) = 2.0 x 10^(-5)

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