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Topic: Ka, Kb values  (Read 27088 times)

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Offline nozo

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Ka, Kb values
« on: August 07, 2006, 07:29:10 AM »
Is my reasoning correct?

Given the following equilibria:

NH4 + H2O <----> NH3 + H3O   Ka = 5.6 x 10-10
HNO2 + H2O <----> NO2- + H3O   Ka = 4.5 x 10-4

The aq solution of NH4NO2 is acidic? basic? neutral?

My reasoning:

NH3+ + HNO2 ---> NH4NO2

I simply look up the Kb value for NH3 and the Ka for HNO2, which are Kb = 1.8 x 10-5,  Ka= 4.5 x 10-4
Since Ka is larger, the soln is acidic

Or am I suppose to solve for Kb given the equilibria above?

Offline Borek

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Re: Ka, Kb values
« Reply #1 on: August 07, 2006, 08:51:02 AM »
You have solution containg weak acid (NH4+) and weak base (NO2-). Compare Ka and Kb to see what is stronger.

Or check this salt pH lecture and look at the equation 13.11
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Offline sdekivit

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Re: Ka, Kb values
« Reply #2 on: August 07, 2006, 10:16:43 AM »
for the comparison of Ka with Kb simply look at the dissociation:

Ka = [H+][NO2-]/[HNO2]

and

Kb = [OH-][NH4+]/[NH3]

When K is high, the denominator is small and thus much dissociation. Thus the equilibrium with the highest K will have a higher dissociation.

Offline nozo

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Re: Ka, Kb values
« Reply #3 on: August 07, 2006, 10:32:08 AM »
Oh okay, I thought NH3+ was the weak base and and HNO2 was the weak acid... so I simply compared their Ka, Kb values... (ugh, acid-base topics are sooo confusing)

So I solved KbKa = Kw

Kb = 1.1 x 10-14 / 4.5 x 10-4 = 2.2 x 10-11 which is less than the weak acid of NH4 = 5.6 x 10-10

Cool... thanks guys ^-^  (ps.. I still hate this topic though :P)

Offline sdekivit

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Re: Ka, Kb values
« Reply #4 on: August 07, 2006, 10:48:18 AM »

I simply look up the Kb value for NH3 and the Ka for HNO2, which are Kb = 1.8 x 10-5,  Ka= 4.5 x 10-4
Since Ka is larger, the soln is acidic


my answer is a reply to these data.

But it's easier to look at the dissociation of the salt itself:

NH4NO2 --> NH4+ + NO2-

And thus we have the equilibria:

1) NH4+ + H2O <--> NH3 + H3O+
2) NO2- + H2O <--> HNO2 + OH-

« Last Edit: August 07, 2006, 11:02:51 AM by sdekivit »

Offline Borek

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Re: Ka, Kb values
« Reply #5 on: August 07, 2006, 10:54:55 AM »
for the comparison of Ka with Kb simply look at the dissociation:

Ka = [H+][NO2-]/[HNO2]

and

Kb = [OH-][NH4+]/[NH3]

When K is high, the denominator is small and thus much dissociation. Thus the equilibrium with the highest K will have a higher dissociation.

Please clarify what you are refering to. Both acid and base present in the solution are conjugated ones, so equations you have posted doesn't say a word about what is happening in the solution.
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Offline sdekivit

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Re: Ka, Kb values
« Reply #6 on: August 07, 2006, 11:00:31 AM »

Please clarify what you are refering to. Both acid and base present in the solution are conjugated ones, so equations you have posted doesn't say a word about what is happening in the solution.

see previous reply.

Nozo did the rest of the problem correctly comparing the different K-values :)
« Last Edit: August 07, 2006, 11:05:40 AM by sdekivit »

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