[makefastice]

It Takes 55.48 ML of NaOH to titrate 25.00 ML of a weak monoprotic acid. After 31.64 ML of NaOH was added The PH was 4.96. What is th Ka and Pka?

[makefastice]

I dont know how to solve pka=ph + log [ha]/[a] without knowing the molarity of the starting acid bace. Will It work to use the ML ratio.
Like : Pka= ph + log( [55.48 ml/ 31.64])?

[mike]

It Takes 55.48 ML of NaOH to titrate 25.00 ML of a weak monoprotic acid. After 31.64 ML of NaOH was added The PH was 4.96. What is th Ka and Pka?

do you mean mL (when you write ML)?

[makefastice]

yes

[makefastice]

ok so at 31.64 ml the base is 57% titrated.   31.64mL/55.48mL= .57
Right?

Then Pka = PH + log(.57)?

[mike]

sounds reasonable to me

[makefastice]

Thanks

[Borek]

No!

Let's put

[HA] + [A^- ] = C

Once 31.64 mL of titrant was added 31.64/55.48 acid was neutralized, so:

[A^- ] = 31.64 C / 55.48

So far so good, but now we need concentration of [HA] - that's what was left:

[HA] = (1 - 31.64 C / 55.48) C

putting into Henderson-Hasselbalch equation gives:

pKa = pH + log ( (1 - 31.64 / 55.48) C / (31.64 C / 55.48))

C cancels out.

One could also take volume changes into account, but luckily finally they cancel out like C.

[mike]

Hey Borek, Ok I am sure you are correct (you seem to be a bit of a champion at concentrations etc )

I just cannot see how you did this calculation

COuld you please explain it too me?

Let's put [HA] + [A- ] = C

is this refering to equation HA <---> H+ + A-

what is C? what is the relevance?

pKa = pH + log ( (1 - 31.64 / 55.48) C / (31.64 C / 55.48))

should it be:

pH = pKa + log[A-]/[HA] ?

I am sorry, I am not being funny I just don't get it?? this is very frustrating!

[Borek]

what is C?

C is total (analytical) concentration of acid. During titration acid will be present in neutralized and undissociated forms, sum of their concentrations will be constant (if you neglect dilution).

When you have added 31.64 mL you have neutralized 31.64/55.48=57% of acid. That means 43% (1-31.64/55.48) is still in HA form. These values should be put into Henderson-Hasselbalch equation.

should it be:

pH = pKa + log[A-]/[HA] ?

It is exactly the same, just solved for pKa (unless there is some error in my math). It can be expressed in many ways, like:

pH = pKa + log (57%/43%)

As you see the difference between your equation and the above is that you forgot about 43% - concentration of undissociated (not neutralized) acid.

You can afford to use percent under logarithm because once again everything cancels out. Only ratio is important.

[mike]

Thank you