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Calculating the pH of a Basic Salt

pH of Acidic Salts link

 

0.50M NaC2H3O2

So what makes this salt basic?

It contains C2H3O2- , the conjugate base of HC2H3O2 (ka=1.8 x 10-5).

HC2H3O2 is a weak acid.

Na+ is the conjugate acid of NaOH and does not form. NaOH is a strong base.

React C2H3O2- with water, remembering bases accept H+.

C2H3O2- (aq)  + H2O(l) ó HC2H3O2 (aq) + OH- (aq)

  

Set up the ICE box
 C2H3O2-(aq)+ H2O(l)

ó HC2H3O2(aq)

+ OH- (aq)
I0.50M

 

------

------
C-X 

+X

+X
E0.50M-X 

+X

+X

Kb=

[HC2H3O2][OH- ]
----------------------------------
[C2H3O2-]

But we have a ka that must be converter to Kb

Kb=

Kw
----------------------------------
Ka

Kb=

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=[OH-]=1.67 x 10-5M 

pOH=-log[OH-]=-log 1.67 x 10-5M 

pOH=4.78

pH=14-4.78

pH=9.22

on to the pH of Acidic Salts link

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