Question

Consider the titration of a 22.0-mL sample of 0.105 M HC2H3O2 with 0.130 M NaOH. (The value of Ka for HC2H3O2 is 1.8×10−5.)

a.Determine the initial pH.= 2.86 pH

b.Determine the volume of added base required to reach the equivalence point.

c.Determine the pH at 6.0 mL of added base.

d.Determine the pH at one-half of the equivalence point. = 4.74 pH

Answer 1

The value of Ka for HC2H3O2 is 1.8×10−5.

a. The initial pH

Ka = 1.8 x 0^-5

Ka = x^2/ 0.105-x

x = 1.37 x 10^-3

x = [H+] =0.00137 M

PH = - log [H+]

PH = - log (0.00137)

PH = 2.86

b. The volume of added base required to reach the equivalence point.

Calculate the moles of acetic acid by using formula n = c v in(L)

Moles of acetic acid = 0.022 x 0.105 =0.00231

Moles of NaOH required = 0.00231

Volume of NaOH = n/c

Volume of NaOH = 0.00231/ 0.130 M

= 0.01776 L

= 17.76 ml

c. Determine the pH at 6.0 mL of added base.

Moles NaOH = 6.00 x 10^-3 L x 0.130 M=0.00078 mol

0.00078 mol of NaOH neutralize 0.00078 mol of acetic acid

Moles acetic acid in excess = 0.00231 - 0.00078 =0.00153

Moles acetate formed = 0.00078

PKa = 4.74

Now use the Henderson-Hasselbalch equation to calculate PH

pH = pKa + log ([A-]/[HA])

[A-] = molar concentration of a conjugate base

[HA] = molar concentration of a undissociated weak acid

PH = 4.74 + log 0.00078/ 0.00153

PH = 4.74 + (-0.29)

PH = 4.45

d. PH at one-half of the equivalence point

At one half equivalence point moles acid = moles acetate

PH = PKa + log ([A-]/[HA])

PH = 4.74 + log 1

PH = 4.74 + 0

PH = 4.74

At one-half of the equivalence point pH = PKa = 4.74