Consider the titration of a 25.0 mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH.

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.

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

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