ka AND kb: The Reverse Reaction?

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capostat

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I read somewhere than Kb is not the reverse reaction of Ka. Why not?

If Ka is the dissociation constant of NH4+ to NH3, then why isnt Kb the hydrolyzation constant of NH3 to NH4+?

Or was the information I read, wrong?

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I read somewhere than Kb is not the reverse reaction of Ka. Why not?

If Ka is the dissociation constant of NH4+ to NH3, then why isnt Kb the hydrolyzation constant of NH3 to NH4+?

Or was the information I read, wrong?


NH4+ + H20 -> NH3 + H30 , and Ka here is [NH3]*[H30]/[NH4+] right?

key point here is that for Ka, the acid dissociation constant, an acid (H30) is produced. now take the reverse reaction of the reaction I just put up there;

NH3 + H30 -> H20 + NH4+ here, no base is produced (OH-). a Kb constant reflects the production of base produced. so the reverse reaction does not show Kb.

the Kb reaction with ammonia in aqueous solution would be
NH3 + H20 -> OH- + NH4+ and Kb would be [OH-]*[NH4+]/[NH3}

does that make sense?
 
NH4+ + H20 -> NH3 + H30 , and Ka here is [NH3]*[H30]/[NH4+] right?

key point here is that for Ka, the acid dissociation constant, an acid (H30) is produced. now take the reverse reaction of the reaction I just put up there;

NH3 + H30 -> H20 + NH4+ here, no base is produced (OH-). a Kb constant reflects the production of base produced. so the reverse reaction does not show Kb.

the Kb reaction with ammonia in aqueous solution would be
NH3 + H20 -> OH- + NH4+ and Kb would be [OH-]*[NH4+]/[NH3}

does that make sense?

ooooh k....

i see...

I have to remember that i have to yield H+ for Ka and OH- for Kb...

there's another question that i might have. i saw it posted on here, i'll try to find...

thanks
 
Alright here it is:

Calculate the pKa of HCO3-. The Ka=4.7x10^-11 and the Kb=2.7*10^-8 for HCO3-.

Choices:
A. 10
B. 8
C. 6
D. 4

Answer: C


How to go about his? I cant imagine that the answer would have been gotten by taking that "-11" and going up a bit to 10, or 10 point something .

HCO3- is the conjugate base to H2CO3.

That Kb=2.7x10^-8 is for this reaction:

So HCO3- + H20--> H2CO3 + OH-

(The clue here is that since Kb is bigger than Ka, we are see that HCO3- is a better base than acid?)

Here's where I'm stuck..

That Ka=4.7x10^-11 is for this reaction:

HCO3- + H20---> CO3- + H3O+


doesnt that mean that the pka for HCO3- is "a little less than 11"?
 
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Alright here it is:

Calculate the pKa of HCO3-. The Ka=4.7x10^-11 and the Kb=2.7*10^-8 for HCO3-.

Choices:
A. 10
B. 8
C. 6
D. 4

Answer: C


How to go about his? I cant imagine that the answer would have been gotten by taking that "-11" and going up a bit to 10, or 10 point something .

HCO3- is the conjugate base to H2CO3.

That Kb=2.7x10^-8 is for this reaction:

So HCO3- + H20--> H2CO3 + OH-

(The clue here is that since Kb is bigger than Ka, we are see that HCO3- is a better base than acid?)

Here's where I'm stuck..

That Ka=4.7x10^-11 is for this reaction:

HCO3- + H20---> CO3- + H3O+


doesnt that mean that the pka for HCO3- is "a little less than 11"?

Something is wrong here...
Ka x Kb = Kw
Kw = 10^(-14)

Kw/Kb = Ka
Ka = (1/2.7)x10^(-6) = approx. 3.3x10^(-7)
pKa = -log(Ka) = between 6 and 7 but closer to 6.

Are you sure u read the question correctly? The Ka x Kb should = Kw for conjugates.
 
Last edited:
I just copied the info from where I saw it, which was here:

http://forums.studentdoctor.net/showthread.php?t=435998&highlight=pka+pkb

you said Ka x Kb = Kw for the same molecule?

does this conflict with pKa + pkb=14 for conjugate pairs?

Its not the same molecule, sorry, its conjugates.

so
Ka x Kb = Kw
pKa + pKb = 14
both for conjugates.

So Kb(HCO3-) x Ka(H2CO3) = Kw
or Ka (HCO3-) x Kb(CO3--) = Kw

The pKa I calculated would be the pKa for H2CO3, which is between 6/7.
The pKa according to this problem for HCO3- would just be -log(4.7x10^-11) which is between 10/11.
 
Its not the same molecule, sorry, its conjugates.

so
Ka x Kb = Kw
pKa + pKb = 14
both for conjugates.

So Kb(HCO3-) x Ka(H2CO3) = Kw
or Ka (HCO3-) x Kb(CO3--) = Kw

The pKa I calculated would be the pKa for H2CO3, which is between 6/7.
The pKa according to this problem for HCO3- would just be -log(4.7x10^-11) which is between 10/11.

ok.

I think i see what you're saying--the problem might have been transcribed wrong.

The goal of the problem should be to find the pka of H2CO3. And you're given pka and pkb of HCO3-.

so just to run this by me again...

kw = ka x kb

and then...

ka=kw/kb

so ka = (1 x 10^-14)/(2.7 x10^-8) = the answer you got,

and then, you take the exponent and subtract a little bit...


so in the end, the ka of HCO3- is to throw you off, i guess.
 
Its not the same molecule, sorry, its conjugates.

so
Ka x Kb = Kw
pKa + pKb = 14
both for conjugates.

So Kb(HCO3-) x Ka(H2CO3) = Kw
or Ka (HCO3-) x Kb(CO3--) = Kw

The pKa I calculated would be the pKa for H2CO3, which is between 6/7.
The pKa according to this problem for HCO3- would just be -log(4.7x10^-11) which is between 10/11.


That's what I got as well, but that doesn't make sense for an acid in solution.
 
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