Respuesta :
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
The values of pH are as follows:
a) 2.72
b) 4.26
c) 4.74
d) 8.65
e) 8.79
f) 12.22
Calculation of pH:
Considered the titration of 100.0 mL of 0.200 M acetic acid (Ka=1.8 x 10^-5) by 0.100 M KOH.
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
Balanced chemical reaction:
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
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