pKa of conjugate acid

[663697]

What is the pKa for the conjugate acid of the solute shown on the curve?

A. 1
B. 2
C. 6
D. 9

I chose D for this, but I think it may have been for the wrong reasons.. I looked at the point where pH = pKa and this was above 7, so I chose 9. However, I have a feeling that this might be the wrong reasoning given that it is a weak base being titrated with a strong acid. Would we have to convert to pOH/pKb here?

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[akimhaneul]

Yeah you can't use the Henderson hasselbach equation we use all the time here because this is a base turning into acid.

When you use the kb to derive an equation similar to the Henderson hasselbach you end up with
Poh = pkb + log (bh+/b)
So you can conclude that pkb is about 8 or 9 like you said which means Pka is 6 or 5.
Therefore c is the answer

Here is a useful information about this. You should read it

Henderson–Hasselbalch equation - Wikipedia


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[desperadoflakes]

You can estimate the equivalence point as being

pEQ ~ ((pOH of pure acid titrant) + (pKb of base)) / 2
In this case, the pEQ looks to be around 4 or 5, and the pH of the pure acid titrant looks to be around 1. We're looking for pKa.

5 ~ (pH of pure acid being titrated + pKb) / 2
5 ~ (1 + pKb) / 2
pKb ~ 9

 

[663697]

You can estimate the equivalence point as being

pEQ ~ ((pOH of pure acid titrant) + (pKb of base)) / 2
In this case, the pEQ looks to be around 4 or 5, and the pH of the pure acid titrant looks to be around 1. We're looking for pKa.

5 ~ (pH of pure acid being titrated + pKb) / 2
5 ~ (1 + pKb) / 2
pKb ~ 9

That would give us pKb, not pKa right?

 

[desperadoflakes]

Sorry, it was late, I made a mistake.

When titrating a weak acid with a strong base, you can estimate the equivalence point with:
pH(eq) = (pKa (initial acid) + pH (titrant base))/2

When titrating a weak base with a strong acid, you can estimate the equivalence point with:
pOH(eq) = (pKb (initial base) + pOH (titrant acid))/2

OR

pH(eq) = (pKa (conjugate acid) + pH (titrant acid))/2

Let's use the second one.

5 = ((pKa) + 1)/2
pKa ~ 9

Really, you don't have to do any math for this problem. You know that the pKb has to be above the equivalence point (when titrating a base). The equivalence point is the point at which you've added an equal number of moles of your strong acid to your initial number of moles of your weak base. That means that, at the equivalence point, ALL of the A- (base) you initially had in solution will be neutralized by the H+ (titrant acid). And you know that the pKa is the point at which HALF of your A- is neutralized (because A- = HA).

When dealing with titrations, you can think of the pKa (of your conjugate acid) as being your HALF-equivalence point, where the vol. of titrant you have added is HALF of what it would be at the equivalence point. When dealing with a titration curve where the y-axis is pH, we are only dealing with pKas. If you're titrating an acid, the half-equivalence point is the pKa of your initial acid. If you're titrating a base, the half-equivalence point is the pKa of your conjugate base.

If the y-axis is pOH, than the graph gets flipped upside down, and all values are really just (14 - what they were before). Now you're dealing with pKb. If you're titrating an acid, it's the pKb of the conjugate base. If you're titrating a base, it's the pKb of the base.

Hope that clears things up!

 

[DaAlienist]

You can estimate the equivalence point as being

pEQ ~ ((pOH of pure acid titrant) + (pKb of base)) / 2
In this case, the pEQ looks to be around 4 or 5, and the pH of the pure acid titrant looks to be around 1. We're looking for pKa.

5 ~ (pH of pure acid being titrated + pKb) / 2
5 ~ (1 + pKb) / 2
pKb ~ 9

No need to do math, just look at the graph. The pH will be equal to the pKa when there are equal amounts of original molecule and its conjugate (since log 1 = 0). This is known as the equivalence point, and will be the steep parts of any titration curve.

Looking at the figure in the Q, the steepest part is just below 7. (9 is impossible, and 1 and 2 are way too low) No math needed at all.