# How to calculate the dissociation constant of a weak acid from the titration with a strong base?

Problem:
A solution of an unknown weak acid of unknown concentration was titrated with a solution of a strong base of unknown concentration. During the titration, the pH after adding $\pu{2.00 mL}$ of the base was 6.912. An additional $\pu{14.00 mL}$ of the base was required to reach the equivalence point. Calculate the Ka of the weak acid.

I have decimated six sheets of paper trying different equations. Combining the two ICE tables you can make for each equation, I can still get no fewer than three unknowns. I'm sure I'm overlooking a super-simple solution but I cannot think of it.

These images are what I have done so far.

• It would be useful if you include at least some of the work you've done in detail than generally which you have done here. This is flagged as a homework question but I vote to leave open for now. Jan 23, 2016 at 11:46

## 1 Answer

This is a strong base weak acid titration. If the weak acid is of the form $\ce{HA}$, then at equivalence point, all the acid has been converted to its conjugate base.

Total volume of base required for equivalence point is $\mathrm{16~ml}$. Let the molarity of base used be $M$. Then, number of milli moles of base for equivalence point is $M\times 16$. This is also the number of milli moles of acid taken.

If only $\mathrm{2~ml}$ of base was used, there will be some acid and some salt of acid and base (A buffer, since the acid is weak and the base is strong). $$pH=pk_a+\log\frac{M\times 2}{M\times 16-M\times 2}=pk_a+\log\frac{2}{14}$$ From this you will get $$6.912=pk_a-0.845$$ $$pk_a=7.757$$