Question

Consider the titration of 100.0 mL of 0.200 M acetic acid () by 0.100 M . Calculate the pH of the resulting solution after the following volumes of have been added. 0.0 mL pH = 50.0 mL pH = 100.0 mL pH = 120.0 mL pH = 200.0 mL pH = 260.0 mL pH =

Answer 1

Answer:

a) 2.72

b) 4.26

c) 4.74

d) 8.65

e) 8.79

f) 12.22

Explanation:

Considered the titration of 100.0 mL of 0.200 M acetic acid (Ka=1.8 x 10^-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.

Step 1: Data given

Volume of 0.200 M acetic acid = 100.0 mL = 0.100 L

Concentration of KOH = 0.100M

Ka of acetic acid = 1.8 * 10^-5

pKa = 4.74

Step 2: The balanced equation

KOH + CH3COOH → CH3COOK + H2O

a) When 0.0 mL is added

pH = - log*(√[HA]*Ka)

pH = -log (√(0.200 * 1.8 x 10^-5)

pH = 2.72

b) When 50.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.0500 L * 0.100 M=0.005  moles

moles acetic acid in excess = 0.0200 - 0.005=0.015  moles

moles acetate = 0.005 moles

pH = pKa + log ([acetate]/[acetic acid])

pH = 4.74 + log 0.005/0.015= 4.26

c) When 100.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.100 L * 0.100 M=0.010 moles

moles acetic acid in excess = 0.0200 - 0.010=0.010 moles

moles acetate = 0.010 moles

CH3COO- + H2O <=> CH3COOH + OH-

pH = 4.74 + log 0.0100/0.0100= 4.74

d) When 120.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.120 L * 0.100 M=0.012 moles

moles acetic acid in excess = 0.0200 - 0.012=0.008  moles

moles acetate = 0.008 moles

total volume = 0.220 L

concentration acetate = 0.008/0.220=0.0364 M

CH3COO- + H2O <=> CH3COOH + OH-

Kb = 5.6 * 10^-10 =x^2/ 0.0364-x

x = [OH-]= 4.51 * 10^-6  M

pOH = 5.35

pH = 8.65

e)  When 200.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.200 L * 0.100 M=0.0200 moles

moles acetic acid in excess = 0.0200 - 0.0200 = 0  moles

moles acetate = 0.020 moles

total volume = 0.300 L

concentration acetate = 0.020/0.300 =0.0667 M

CH3COO- + H2O <=> CH3COOH + OH-

Kb = 5.6 * 10^-10 =x^2/ 0.0667-x

x = [OH-]=6.11 * 10^-6  M

pOH = 5.21

pH = 8.79

f)  When 260.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.260 L * 0.100 M=0.0260 moles

moles OH- in excess = 0.0260 - 0.0200 = 0.0060 moles

total volume = 0.360 L

[OH-] = 0.0060 moles / 0.360 L

[OH-]=  0.0167

pOH = -log [0.0167]

pOH = 1.78

pH = 14- 1.78 = 12.22