TPRSW General Chemistry Passage 51 #5

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

faith hopelove

God Fearing
10+ Year Member
Joined
Jun 14, 2012
Messages
99
Reaction score
6
The question was:
What is the pka for a conjugate acid of the solute in solution X?
a. 1
b. 2
c. 3
d. 9

Now the passage stated that the pH of Solution X is 10.7 and the titration graph reflects this (monoprotic solution).

The explanation in the back of the book gave a long drawn out solution stating that we need to change the pH to pOH to find the pkb to get the pka. But I want to know is if we already know that pH of the solution and pH=pka at the half mark to the equivalence point, do we really need to be going from pH→pOH→pkb→pka?

Members don't see this ad.
 
Are they showing you that pH = 10.7 at half-equivalence or equivalence pt? I'm assuming that X is a base being titrated with an acid, and the graph shows pH = 10.7 at equivalence point, correct me if I'm wrong as I can't 'see' the question yet.
 
Here is a picture of what I was trying to describe.
 

Attachments

  • FullSizeRender.jpg
    FullSizeRender.jpg
    78 KB · Views: 41
You don't need to be going from pH to pOH.

If titrating a weak acid with a strong base:

At half eq. point:
pH = pKa of the weak acid.
pOH = pKb of the conjugate base.

If titrating a weak base with a strong acid:

At half eq. point:
pOH = pKb of weak base.
pH = pKa of conjugate acid.
 
At half equivalence the pH is 10.7 but since we have a base, X, not an acid, we do need to find the pkb from pOH.

For titration of a weak base with a strong acid, (as Odi says above pOH = pKb). Since we are given the pH not pOH here.

As the solution states we have to convert pH to pOH and then find the pKb from pOH. We can then find the pKa of the conjugate acid of X, which is what the question asks for, not the pKa of X.
 
Top