DAT Destroyer 2012 GenChem #147

[monsieurwise]

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Hi everyone,
I have a very basic acid/base problem here but I have no idea why I am still confused with this:
What is the pH of a .05 M NH3 solution if Kb is 1.8 x 10^-5?

I tried to write as this:
NH4+ ----> NH + H+

And I convert Kb to Ka, Ka=5.6 x 10^-5

So I have (X^2)/(.05) = Ka

And I do calculations to find X=[H+] and then find pH.

And my answer came out wrong. Is it because [NH4+] is not .05M?

Thank you so much guys!!

 

[Meat Gyver]

NH3 + H2O => NH4+ + OH-

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

Everything is in a 1:1:1 mole ratio, so when you solve for X, you can get the [OH-] to solve pOH.

Kb = [x][x]
[.05M]

(1.8 x 10^-5) (.05M) = x^2

X ^2= 9 x 10 -7 ------------>make it into 90 x 10^-8 to get a better square

Square root that: (you know 90 is between the square root of 9 (81) and 10 (100)...its closer to 100 so its larger than 9 (estimate around 9.5)

x = 9.5 x 10^ -4 = [OH-]

pOH = -log (9.5 x 10 -4) = estimate to 3.1

pOH + pH = 14

3.1 + pH = 14

pH = about 11

Is that the right answer?

 

[Meat Gyver]

Just note:

Ka is the dissociation constant for an acid in water:

HA + H2O <==> H3O+ + A- Ka = ([H3O+][A-]) / [HA]

Kb is the dissociation constant for a base in water:
B + H2O <==> HB + OH- Kb = ([HB][OH-]) /

NH3, despite being a weak base, is still a base and won't give off H+ ions when it dissociates in water. So converting Kb to Ka won't help.

So if given Kb of any base and asked to find pH, you have to first solve for [OH-], because that's all that you can take out of Kb = ([HB][OH-]) / . After that calculate pOH and with that you can figure out pH.

 

[monsieurwise]

Ah ok thank you! So what does the Kb of an acid like HCl means? Is it the Kb of Cl-?
Thank you so much for such a detailed answer!! =))

 

[Meat Gyver]

HCl is a strong acid so Ka's and Kb's don't apply to it. HCl has Ka of 1.3 * 10^6 (much larger than 1) so its conjugate base Cl- is neutral in terms of basicity. In other words, its really stable by itself and wont accept a H+ from water.

However, if you have a weak acid (Ka), it will have a conjugate base (Kb)

Using Ka*Kb = Kw, if you know the kA for a weak acid, you can calculate the Kb for its conjugate base.