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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nozo on August 07, 2006, 07:29:10 AM

Title: Ka, Kb values
Post by: nozo 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?

Title: Re: Ka, Kb values
Post by: Borek 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 (http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified) and look at the equation 13.11

Title: Re: Ka, Kb values
Post by: sdekivit on August 07, 2006, 10:16:43 AM

for the comparison of K_a with K_b simply look at the dissociation:

K_a = [H⁺][NO₂^-]/[HNO₂]

and

K_b = [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.

Title: Re: Ka, Kb values
Post by: nozo 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)

Title: Re: Ka, Kb values
Post by: sdekivit on August 07, 2006, 10:48:18 AM

Quote from: nozo on August 07, 2006, 07:29:10 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^-

Title: Re: Ka, Kb values
Post by: Borek on August 07, 2006, 10:54:55 AM

Quote from: sdekivit on August 07, 2006, 10:16:43 AM

for the comparison of K_a with K_b simply look at the dissociation:

K_a = [H⁺][NO₂^-]/[HNO₂]

and

K_b = [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.

Title: Re: Ka, Kb values
Post by: sdekivit on August 07, 2006, 11:00:31 AM

Quote from: Borek on August 07, 2006, 10:54:55 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 :)