SCH 4U - Weak Acids

Weak Acids

Most acids are weak acids and tend to dissociate partially in solution.

HA(aq)  ↔  H⁺(aq ) +  A^-(aq)

K_a = [H⁺][A^-]

        ----------

          [HA]        (the larger the value of K_a, the stronger the acid)

Notice that the subscript on the equilibrium constant is now “a” to indicate it pertains to an acid.   There are tables of K_a values:

For acids that have more than one ionizable hydrogen ion, there are multiple K_a values (K_a1, K_a2, K_a3).  So, for an acid like H₃PO₄, it would work like this, since polyprotic acids will dissociate one proton at a time:

H₃PO₄(aq)   ↔  H⁺(aq)  +  H₂PO₄^-(aq)     Ka1 = 7.5 x 10^-3

H₂PO₄^-(aq)   ↔  H⁺(aq)  +  HPO₄^2-(aq)    Ka2 = 6.2 x 10^-8

HPO₄^2-(aq)   ↔  H⁺(aq)  +  PO₄^3-(aq)       Ka3 = 4.2 x 10^-13

 

👉 Be aware that values will be slightly different from table to table.  You may find in my answer keys that the value I have used is slightly different from what you have used.  As long as your technique matches mine, it's all good. 👈

We will incorporate K_a along with our best bud, the ICE table, to do some rad calculations.  So, let’s get this party started:

Calculating [H⁺] & pH of a Weak Acid, Given K_a     

The following question is solved in a similar manner to questions we have done in the past, so I'm just going to let the answer speak for itself.  The only difference is, at the end, pH is calculated.

Example:  Calculate the pH of a 0.30 M solution of acetic acid.

Answer:     

               H₃CCOOH(aq)  ↔  H⁺(aq)  +  CH₃COO^-(aq)          K_a = 1.8 x 10^-5

in [  ]               0.30                    0                    0

Δ [  ]                 -x                     +x                  +x

eq [  ]           0.30 – x                 x                     x

           K_a = [H⁺][CH₃COO^-]/[CH₃COOH]                    pH = -log[H⁺]

1.8  x 10^-5 = x²/(0.30 – x)                                                   = -log(2.3 x 10^-3)

1.8 x 10^-5 = x²/0.30                                                           = 2.63

        x = 2.3 x 10^-3

Finding K_a, Given Concentration & pH

This is also similar to questions done in the beginning portion of the equilibrium unit.  However, in this case, before we can fill in the ICE table, we must use the pH to determine the [H⁺].  This value is placed in the eq section of the ICE table under H⁺ and the remainder of the table is filled in logically from there.  

Example: A 0.182 M solution of an acid, HX, has a pH of 3.21.  What is the K_a of the acid?

Answer:

                  HX(aq)        ↔        H⁺(aq)     +      X^-(aq)         

in [  ]           0.182                         0                    0

Δ [  ]      -6.17 x 10^-4            +6.17 x 10^-4     +6.17 x 10^-4

eq [  ]          0.181                   6.17 x 10^-4          6.17 x 10^-4   

pH = -log[H⁺]                      K_a = [H⁺][X^-]/[HX]

-3.21 = log[H⁺]                         = (6.17 x 10^-4)²/0.181

[H⁺] = 6.17 x 10^-4 M                 = 2.10 x 10^-6

Homework # 14-15, 17