5
$\begingroup$

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. enter image description here enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – M.A.R. ಠ_ಠ Jan 23 '16 at 11:46
6
$\begingroup$

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 $$6.912=pk_a-0.845$$ $$pk_a=7.757$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.