Respuesta :
Answer:
1) pH = 4.51
2) pH = 4.87.
3) pH = 12.32
Explanation:
1) the Ka of propanoic acid is 1.34 X 10⁻⁵
Therefore pKa = 4.87
When we add 5 mL of 0.300 M NaOH the moles of base added is
moles = molarity X volume
moles = 0.300 X 5mL = 1.5 mmoles
moles of acid present = molarity X volume = 0.165 X 30.0 = 4.95 mmoles
on addition of 1.5 mmoles of base the moles of acid neutralized = 1.5mmole
This will result in formation of salt of the acid
the moles of salt formed = 1.5 mmoles
the moles of acid left = 4.95 - 1.5 = 3.45 mmol
this acid and its salt mixture results in formation of a buffer
the pH of buffer is calculated as:
pH = pKa + log [salt] / [acid]
pH = 4.87 + log [1.5/3.45] = 4.51
2) at half equivalence point the moles of acid becomes equal to moles of salt formed thus the pH of solution will become equal to the pKa of acid
pH = 4.87.
3) the moles of based added due to addition of 20.0 mL = molarity X volume
moles = 0.300 X 20 = 6mmol
This will completely neutralize the acid (4.95 mmol)
after neutralization the moles of base left = 6-4.95 = 1.05 mmol
Total volume of solution = volume of acid + volume of base =30+20=50
concentration of hydroxide ion (due to excess base) = [tex]\frac{mmoles}{volume(mL)}[/tex]
[OH⁻]=0.021
pOH = -log[OH⁻]=1.68
pH = 14-pOH = 12.32
The titration results in the neutralization reaction with the addition of acid to the base. The pH with the addition of 5 ml of the base is 4.51. The pH at half equivalence is 4.87. The pH with 20 mL base is 12.32.
What is pH?
The pH can be defined as the concentration of hydrogen ions in the solution.
The moles of acid in the solution is:
[tex]\rm Moles=Molarity\;\times\;Volume(L)[/tex]
- Moles of propionic acid are:
[tex]\rm Moles\;propionic \;acid=0.165\;\times\;0.03\;L\\Moles\;propionic\;acid=4.95\;mmoles[/tex]
The moles of base, KOH in 5 mL will be:
[tex]\rm Moles\;KOH=0.3\;M\;\times\;0.0005\;L\\Moles\;KOH=1.5\;mmoles[/tex]
The moles of acid left after neutralization is:
[tex]\rm Moles\;acid\;left=4.95-1.15\;mmoles\\Moles\;acid\;left=3.45\;mmoles[/tex]
The pH of the solution can be given as:
[tex]\rm pH =pKa+log\dfrac{salt}{acid}\\ pH=4.87+log\dfrac{1.5}{3.45}\\pH=4.51[/tex]
- The pH of the solution after the addition of 5 ml base is 4.51.
The pH at half equivalence point will be equivalent to the pKa, as the moles of salt is equivalent to the moles of acid.
Thus, the pH at the half equivalence point is 4.87.
- The pH after the addition of 20 mL base is given as:
The moles of base added is:
[tex]\rm Moles\;KOH=0.3\;M\;\times\;0.02\;L\\Moles\;KOH=6\;mmol[/tex]
The acid will be completely neutralized with the formation of 4.95 mmol of salt. The base left in the reaction will be:
[tex]\rm Moles\;KOH\;left=6\;mmol-4.95\;mmol\\Moles\;KOH\;left=1.05\;mmol[/tex]
The final volume of the solution will be 50 mL. The molarity of the KOH in the solution will be:
[tex]\rm Molariy\;OH^-=\dfrac{mmoles}{volume}\\ Molarity\;OH^-=\dfrac{1.05}{50} \\Molarity\;OH^-=0.021[/tex]
The pOH of the solution is given as:
[tex]\rm pOH=-log[OH^-]\\pOH=-log(0.021)\\pOH=1.68[/tex]
The pH of the solution will be:
[tex]\rm pH=14-pOH\\pH=14-1.68\\pH=12.32[/tex]
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