SCH 4U - Buffers

Buffers

A buffer is a solution that resists changes in pH.  In the buffer, an acid neutralizes added OH^- & a base neutralizes added H⁺.  However, the acid and base must not neutralize each other, so we use a weak acid-weak base conjugate pair like HAc-Ac^-* or NH₄⁺-NH₃.  We can create a buffer of any pH.

HA(aq)   ↔  H⁺(aq)  +  A^-(aq)

K_a = [H⁺][A^-]

        ----------

         [HA]

Buffering Action

If acid (H⁺) is added to a buffer, initially the [H⁺] increases, but the reaction shifts left toward reactants - this decreases the [H⁺] and the affect is very little change in pH.

If base (OH^-) is added to a buffer, initially the [OH^-] increases, but the reaction shifts right toward products, increasing the amount of hydrogen ion present - this decreases the [OH^-] and the affect is very little change in pH.

Buffer Calculations

The calculations for buffers can be done using an ICE table, as shown in the example directly below.  However, an easier, more time-efficient calculation can be done by employing the Henderson-Hasselbalch equation (which is derived after this example). 

Example:  Calculate the pH of a buffer made by adding equal volumes of 0.20 M sodium fluoride to 0.10 M hydrofluoric acid.

     HF(aq)     ↔     H⁺(aq)     +     F^-(aq)

I          0.10                    0                  0.20

C          -x                     +x                   +x

E      0.10 – x                +x              0.20 + x

            K_a = [H⁺][F^-]/[HF]

 6.7 x 10^-4 = x(0.20 + x)/(0.10 –x)

               x = 6.7 x 10^-4/((0.20)(0.10))

               x = 3.4 x 10^-4

pH = -log[H⁺]

pH = -log(3.4 x 10^-4)

pH = 3.47

Henderson-Hasselbalch Equation

To derive the Henderson-Hasselbalch equation, start with the general K_a expression from the start of this lesson:

K_a = [H⁺][A^-]

        ---------

          [HA]

Rearranging slightly:

K_a = [H⁺] x  ([A^-]/[HA])

Take the log of both sides:

log K_a = log[H⁺] + log([A^-]/[HA])

Multiply each term by -1:                                 

-log K_a = -log[H⁺] - log([A^-]/[HA])

In the same way that pH = -log[H⁺], pK_a = -logK_a

pK_a = pH - log([A^-]/[HA])

Rearranging slightly, gives the Henderson-Hasselbalch equation:

pH = pK_a + log([A^-]/[HA])

Revisting the example from above, using the Henderson-Hasselbalch equation gives the same value, with a lot less fuss and muss:

  

pKa = - logKa = -log (6.7 x 10^-4) = 3.17

 

pH = pK_a + log([X^-]/[HX])

pH = 3.17 + log(0.20/0.10)

pH = 3.17 + 0.30

pH = 3.47

Making Buffers

Example: Create a buffer of pH 4.80, using an acid of 0.30 M.

Answer:

Find an acid for which the pH ≃ pK_a:

The buffer pH for which we are aiming is 4.80, so look on the K_a table for a K_a with an exponent of -4 or -5.  There are a few possibilities, but since acetic acid is common, cheap and fairly safe (all qualities that are favourable), let’s use acetic acid, which has K_a = 1.8 x 10^-5.

Determine the pK_a for the chosen acid.

pK_a = -logK_a

       = -log(1.8 x 10^-5)

       = 4.74

Substitute the data into the Henderson-Hasselbalch equation:

  pH = pK_a + log([Ac^-]/[HAc])           

4.80 = 4.74 + log([Ac^-]/[HAc])

4.80 = 4.74 + log([Ac^-]/0.30)

 

Simplify:

0.06 = log([Ac^-]/0.30)

Take the anti-log of both sides of the equation:     

10^0.06 = [Ac^-]/0.30

1.15 = [Ac^-]/0.30

[Ac^-] = 1.15(0.30)

[Ac^-] = 0.35 M

         

Thus, to make the buffer, add equal volumes of 0.30 M HAc and 0.35 M Ac^- (in the form of something like NaAc or KAc).

* I will often use short forms to stand in for longer formulae.  For instance, here I use HAc to represent HC₂H₃O₂ (acetic acid).  Feel free to follow suit and use short forms like I do or use the actual formula – your choice.

Homework # 34-37

Acids & Bases Review can be found here.