Consider the titration of a 25.0 mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH.
ID: 882594 • Letter: C
Question
Consider the titration of a 25.0 mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH. Determine each of the following.
the initial pH
Express your answer using two decimal places.
2.87
B)
Part B
the volume of added base required to reach the equivalence point in mL?
Part C
the pH at 4.00 mL of added base
Express your answer using two decimal places.
Part D
the pH at one-half of the equivalence point
Express your answer using two decimal places.
Part E
the pH at the equivalence point
Express your answer using two decimal places.
pH =2.87
Explanation / Answer
B) For complete neutralization the moles of base required will be equal to moles of acid present
Moles of acid = molarity of acetic acid X volume of acid = 0.1 X 25 = 2.5 millimoles
so moles of base required = 2.5 millimoles
Volume of base = moles of base / molarity = 2.5 / 0.12 = 20.833 mL
C) When 4mL of base is added then moles of base added = volume of base X molarity of base = 4 X 0.12 = 0.48 millmoles
Reaction between acid and base will be
CH3COOH + NaOH --> CH3COO-Na+ + H2O
So 0.48 millimoles of base will neutralize 0.48 millmoles of acid and will produce 0.48 millmoles of salt
The moles of acid left = 2.5-0.48 = 2.02 millmoles
The solution will behave as buffer whose pH can be defined as (pKa of acetic acid = 4.75)
pH = pKa + log [salt] / [Acid] = 4.75 + log 0.48 / 2.02 = 4.75 - 0.624 = 4.126
D) At one half equivalene point
[salt] = [Acid]
so pH becomes equal to pKa
So pH = 4.75
E) pH at equivalence point
At equivalence point the moles of acid neutralized = moles of base added = moles of acetate ions formed = 2.5 millimoles ( as calcualted above)
Concentration of acetate = moles of acetate / total volume = 2.5 / 20.83 + 25 = 0.0545 molar
The acetate ion will hydrolysed as
CH3COO- + H2O --> CH3COOH + OH-
molarity before equilibrium 0.0545 0 0
Change -x x x
molarity at equilibrium 0.0545-x x x
he equilibrium constant expression is
Kb = [OH-(aq)][HC2H3O2(aq)]/[C2H3O2-(aq)] = x2/(0.0545 - x)
Kb = Kw / Ka = 10^-14 / 1.8 X 10^-5 = 5.6 X 10^-10
We can ignore x in denominator
so Kb = x2/0.0545
x2 = 5.6 X 10^-10 X 0.0545
x = 0.552 X 10^-5
[OH-] = 0.552 X 10^-5
So pOH = 5.25
pH = 8.75