1. Using the values below , what is the ka of your unkniwn acid? pH = pKa + log
ID: 1018672 • Letter: 1
Question
1. Using the values below , what is the ka of your unkniwn acid?
pH = pKa + log ([A-])/([HA])
a) From the graph, determine the volume of NaOH required to reach the equivalence point of the titration.
Vol. NaOH at eq. pt. __15.3__
b) Calculate the volume of NaOH required to reach the half equivalence point. Vol. NaOH at half eq. pt. __7.5__
c) From the graph, what is the pH at the halk equivalence point? pH at half eq. pt. ___3.5___
Determination of unknown acid 12 10 8 10 12 14 16 18 Volume of NaOH (mL)Explanation / Answer
pH at half equivalence point is 3.5 will not be right if the total NaOH required for titration is 15.3 it will be 7.65 mL, and pH at this point will be more like 3.7 I would say.
I will do the calculation for both and you can choose.
From the equation
pH = pKa + log base/acid
It is clear that at half equivalence point
base= acid
So log base/acid = log 1 which is 0
so pH = pKa
SO pKa of this acid is either 3.5 or 3.7
Ka = 10-pKa
Ka = 10-3.5 or 10-3.7
Ka = 3.16 x 10-4 or 1.99 x 10-4
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