The Math of Weak Titration

 

Let’s work through our weak titration problem. Here it is again: 

50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

Before the Titration

The first part of the question is just an equilibrium problem. Here is our ICE table:

That leads to the following math:

\(2.56x10^{-5} = \frac{(x) \cdot (x)}{(2.00 - x)}\)

Solving for x, gives us: 

\(x = [H_3O^+] = 0.00714 M\). So, pH = -log(0.00714) = 2.15

It is worth noting that this is notably higher than the starting point of the strong titration.

Between the beginning and equivalence

Now we’ll start the real work of the problem. For each step of the problem, we’ll start with the same table set-up we used before to find the moles of the acid (HA) and the conjugate base \((A^{-1})\). All of the problems leading up to the equivalence point have the same structure.

First we work through the reaction of the acid with the strong base:

then, we set up the ICE table

and solve it

\(2.56x10^{-5} = \frac{(x) \cdot (0.0196 + x)}{(1.94 - x)}\)

Solving for x, gives us:

\(x = [H_3O^+] = 0.00227 M\). So, pH = -log(0.00714) = 2.64

Here's the second row:

\(2.56x10^{-5} = \frac{(x) \cdot (0.167 + x)}{(1.50 - x)}\)

\(x = [H_3O^+] = 2.30x10^{-4} M\). So, pH = -log(0.00714) = 3.64

and the third row:

\(2.56x10^{-5} = \frac{(x) \cdot (0.50 + x)}{(0.50 - x)}\)

\(x = [H_3O^+] = 2.56x10^{-5} M\). So, pH = -log(0.00714) = 4.59

and the fourth row:

\(2.56x10^{-5} = \frac{(x) \cdot (0.66 + x)}{(0.0135 - x)}\)

\(x = [H_3O^+] = 5.24x10^{-7} M\). So, pH = -log(0.00714) = 6.28

The equivalence point

Our next point is the equivalence point, but that presents a problem. We cannot do the same math (using \(K_a\)) because there is NO HA present after the acid base reaction. In fact the only thing present (other than water) after the acid/base reaction is the conjugate base \((A^{-1})\)

When that base is present in water, the following reaction can occur:

\(A^{-1} + H_2O \rightleftharpoons HA + OH^{-1}\)

In order to use this reaction, we need a K value. We know that \(K_w = K_a \cdot K_b\), so solving for \(K_b\) gives us the following:

\(K_b = \frac{K_w}{K_a} = \frac{1.00x10^{-14}}{2.56x10^{-5}} = 3.91x10^{-10}\)

We can create a new ICE table with our new reaction:       

We can now solve this for x:

\(3.91x10^{-10}=\frac{(x) \cdot (x)}{(.667-x)}\) which gives \(1.61x10^{-5}\)

We need to remember that \(x=[OH^{-1}]\) not \([H_3O^+]\), so -log(x) will give pOH, rather than pH.

\(-log(1.61x10^{-10}) = 4.79\)

Then, since pH + pOH = 14, pH = 14 - pOH. So pH = 9.21. This is distinctly ABOVE 7, which means that the equivalence point is NOT neutral.

Beyond equivalence

Past the equivalence point, the problem changes again. Since we have run out of acid (at the equivalence point), any additional amount of base added after that will remain in solution. Since this base is strong, the pH will depend ONLY on this excess base.

\(pOH = -log(\frac{moles~excess~strong~base}{total~volume}) = -log(\frac{0.020 moles}{0.170 L}) = 0.93\)

Therefore the pH (14 - pOH) = 13.07

If this seems like a bit of a “cheat” at the end of a difficult problem, it is. The “truth” is that you should really do another ICE table, using \(K_b\) and the new values.

However, because \(K_b\) is very small (\(10^{-10}\)) and because there is already an amount of product present, the “x” will be very small. Since the \([OH^{-1}]\) = 0.118 + x, if that “x” value is very small, it will not have a mathematically significant effect on the final calculated pOH. 

Solve it yourself if you aren’t sure!

A note about buffers and weak titration

During a weak titration, a buffer solution is present for part of the process. Specifically, between the time you have added about 10% of the base needed to reach equivalence until about  90% of the equivalence volume, the solution behaves as a buffer. That means for those points, you could use the Henderson/Hasselbalch equation. In the problem above, since 100 mL of base were required to reach equivalence, the H/H equation to find the pH between 10 mL and 90 mL.