Thanks Bri MT!
So, just to make sure, in the first instance it is talking about the pKa of the indicator, and in the second instance, it is talking about the pKa of the unknown solution?
I really want to make sure I have a firm grip of the titration concepts, so are there any worksheets with simple and challenging titration problems I can do (preferably with worked solutions, if not I can confirm my answers with you)?
Thanks so much again Bri, appreciate it.
I've edited my above post for clarity since titration jargon can be confusing but yeah.
Let's say, for example, that you have NaOH in your burette (titrant, known concentration) and you're trying to figure out what the concentration of some acetic/ethanoic acid (analyte) is. To find the concentration of the acetic acid you want to identify the volume of NaOH required to reach the equivalence point, where the stoichiometric ratio is met. To visually approximate this, an indicator is added which is a weak acid/base that will quickly change which conjugate is present in a greater quantity and, since each conjugate is a different colour, will change colour at that point. When the indicator transitions from one colour to another, there is the same amount of each conjugate present and the pKa (of the indicator) = pH (of the solution). In this example, your indicator might be something like phenolphthalein
Before the equivalence point is reached, there will be a stage where you've added sufficient titrant (in this example NaOH) that half of the analyte (in this case acetic acid) is protonated and half is deprotonated - in other words, half is the conjugate acid and half is the conjugate base. Thus, we are looking at another situation where pKa =pH but this time it's the pKa of our acetic acid since that's what has a 50/50 balance with conjugate base & conjugate acid. For half of our acetic acid to be deprotonated, the volume of NaOH required is half that needed to fully deprotonate our acetic acid in accordance with the equation NaOH + CH3COOH -> H2O + CH3OONa . I.e. the v(NaOH) is half of what v(NaOH) is at the equivalence point.
Remembering that the stoichiometric ratio is met at the equivalence point (which we approximate with the end point, where the indicator changes colour), this is why pH = pKa at the half equivalence point for the acetic acid and is at the equivalence point for the indicator.
Hope this clarifies things!
I don't think I have a worksheet like that on hand but I'll look around and see if I can find any good question sets like that for you. If you haven't already, you might find it useful to do the maths for why pH = pKa when each conjugate is present in the same amount to solidify your understanding.