Hi, would anyone be able to explain dataset 1 - item 3. Im not sure how pKa = 14 - 9.2 as i thought at the half-equivalence point pKa = pH
TIA
Question: https://www.qcaa.qld.edu.au/downloads/senior-qce/sciences/snr_chemistry_19_ia1_smple_ass_inst.pdf
Answers: https://www.qcaa.qld.edu.au/downloads/senior-qce/sciences/snr_chemistry_19_ia1_smple_m_scheme.pdf
Hey!
The first thing that draws my attention here is is:
Ammonia is a
base and is the substance of unknown concentration (analyte)
We are adding to this our HCl, an acid and our substance of known concentration (titrant)
For all of this I'm assuming 25 C as per usual in QCE chem pH questions
Remember we started with a base and we're adding acid, at this stage we are still on the basic side. At the half-equivalence point we have that half of the base has reacted with the HCl, so pKb = pOH by the exact same logic that gets you to the more commonly discussed pH = pKa.
Remembering that: pKa + pKb = 14 = pH + pOH, we then just do a quick conversion to get the desired value.
We can think of it like this:
An ammonium ion (right hand side of the equation) has a pKa value of roughly 9.2. We want to get the reverse-direction value, ammonia's one, so we do 14 - pk[ammonium] = 14 - 9.2.
We can also think of it like this:
We have pH of 9.2. We want pOH since we're looking at the constant for a base. 14 - pH = 14 - 9.2
TLDR; QCAA used pKa even though they were asking you about a base, which makes things confusing.
Let me know if this clears things up