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Calculating the pH of an Acidic Salt

pH of Basic Salts link

 

0.50M NH4Cl

So what makes this salt acidic?

It contains NH4+ the conjugate acid of NH3 (kb=1.8 x 10-5). NH3 is a weak base.

Cl- is the conjugate base of HCl and does not form. HCl is a strong acid.

 NH4+ will react with water, remembering acids donate H+.

NH4+(aq)  + H2O(l) NH3(aq) + H3O+(aq)

  

Set up the ICE box
  NH4+(aq) + H2O(l)

NH3(aq)

+ H3O+(aq)
I 0.50M

 

------

------
C -X  

+X

+X
E 0.50M-X  

+X

+X

Ka=

[NH3][H3O+]
----------------------------------
[NH4+]

But we have a kb that must be converter to Ka

Ka=

Kw
----------------------------------
Kb

Ka=

10-14

=5.56 x 10-10

----------------------------------
1.8 x 10-5

Now we can plug in our equilibrium values

5.56 x 10-10=

[X][X]
----------------------------------
[0.50-X]

X=[H3O+]=1.67 x 10-5M   [H3O+] is [H+]

pH=-log[H+]=-log1.67 x 10-5M

pH=4.78

on to the pH of Basic Salts link

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