Respuesta :
Answer:
a.pH = 2.51
b.pH = 4.01
c.pH = 5.60
d.pH = 8.35
e.pH = 11.09
Explanation:
Ka = 9.8 x 10^-5
Ka = 4.01
a) no KOH added :
pH = 1/2 (pKa - log C)
= 1/2 (4.01 - log 0.1)
pH = 2.51
b)
millimoles of acid = 40 x 0.1 = 4
millimoles of KOH = 20 x 0.1 = 2
this is half equivalence point :
here pH = pKa
pH = 4.01
c)
millimoles of KOH = 39 x 0.1 = 3.9
acid + KOH -------------> salt + H2O
4 3.9 0
0.1 0 3.9
pH = pKa + log [salt / acid]
= 4.01 + log [3.9 / 0.1]
= 5.60
pH = 5.60
d)
Millimoles of KOH = 40 x 0.1 = 4
Here, salt remains.
salt concentration = 4 / 40 + 40 = 0.05 M
pH = 7 + 1/2 (pKa + log C)
= 7 + 1/2 (4.01 + log 0.05)
= 8.35
pH = 8.35
e)
pH = 11.09
The pH of titration with no KOH is 2.51, with 20 mL KOH is 4.01, with 39 mL KOH is 5.60, with 40 mL KOH is 8.35, and with 41 mL KOH is 8.357.
What is titration?
The titration is the qualitative and quantitative reaction for the determination of one compound, with the addition of another compound.
The pH of the solution in the titration is given as:
- With no KOH:
Since, no KOH is added, the pH has been contributed by barbituric acid:
[tex]\rm pH=\dfrac{1}{2} (pKa-log \;C)\\pH=\dfrac{1}{2} (-log(9.8\;\times\;10^-^5)-log(0.1))\\pH=2.51[/tex]
The pH of the solution is 2.51.
- With 20 mL KOH:
The moles of acid in the reaction are:
[tex]\rm Barbituric\;acid=Molarity\;\times\;volume\\ Barbituric\;acid=0.1\;\times\;0.040\;L\\ Barbituric\;acid=4\;mmol[/tex]
The moles of base has been:
[tex]\rm KOH=Molarity\;\times\;volume\\ KOH=0.1\;\times\;0.020\;L\\ KOH=2\;mmol[/tex]
The reaction has been the half reaction, and the pH is given as:
[tex]\rm pH=pKa\\pH=-log\;( 9.8 \times 10^-5)\\pH=4.01[/tex]
The pH of the solution is 4.01.
- With 39 mL KOH:
The amount of KOH has been equivalent to the salt formed. The salt formed is 3.9 M. The pH is given as:
[tex]\rm pH=pKa\;+\;log\;\dfrac{salt}{acid}\\ pH=4.01\;+\;log\;\dfrac{3.9}{0.1}\\ pH=5.60[/tex]
The pH of the solution is 5.60.
- With 40 mL KOH:
The moles of KOH added is 4 mmol. The salt remained in the reaction will be 0.05 M. The pH of the solution is given as:
[tex]\rm pH=7\;+\;\dfrac{1}{2} (pKa+log\;C)\\pH=7+\;\dfrac{1}{2} (4.01+log\;0.05)\\pH=8.35[/tex]
The pH of the solution is 8.35.
- With 41 mL KOH solution;
The moles of KOH added will be 4.1 mmol.
The salt remained in the reaction is 0.0506 M.
The pH of the solution will be:
[tex]\rm pH=7\;+\;\dfrac{1}{2} (pKa+log\;C)\\pH=7+\;\dfrac{1}{2} (4.01+log\;0.0506)\\pH=8.357[/tex]
The pH of the solution is 8.357.
Learn more about pH, here:
https://brainly.com/question/15289741
