Finding Ka from Kw/Kb  [ENDORSED]

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Rhiannon Imbeah 2I
Posts: 30
Joined: Fri Sep 25, 2015 3:00 am

Finding Ka from Kw/Kb

Postby Rhiannon Imbeah 2I » Tue Jul 26, 2016 7:52 pm

In an example in the course reader for calculating the pH of the 0.15M NH4Cl, the given Kb for NH3 = 1.8x10^-5. There is a step that calculated an amount for Ka by dividing Kw by Kb:

Ka Kb=Kw=10^-14

Ka =Kw/Kb= 5.6x 10^-10.

I am not sure how this answer for Ka was calculated.

Julie Nguyen 1B
Posts: 6
Joined: Fri Sep 25, 2015 3:00 am

Re: Finding Ka from Kw/Kb  [ENDORSED]

Postby Julie Nguyen 1B » Tue Jul 26, 2016 10:33 pm

The equation you would use here is Ka x Kb = Kw. Given the Kb and Kw, you can find the Ka by plugging in values with the following:

Ka = ?
Kb = 1.8x10/^-5
Kw= 10^-14

Proceed to use the equation: Ka x Kb = Kw
Ka x (1.8x10^-5) = 10^-14.
~ divide both sides by 1.8x10^-5 ~
Ka = 5.6x10^-10

And there you have it!

Rhiannon Imbeah 2I
Posts: 30
Joined: Fri Sep 25, 2015 3:00 am

Re: Finding Ka from Kw/Kb

Postby Rhiannon Imbeah 2I » Tue Jul 26, 2016 11:09 pm

Oh yes! Thank you very much


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