Titration of 20.00mL of HC2H3O2(aq) of M=.500M and Ka=1.8x10^-5 with .500M KOH.

Question

Titration of 20.00mL of HC2H3O2(aq) of M=.500M and Ka=1.8x10^-5 with .500M KOH.
Determine the pH before any OH- is added? Also Determine the pH after the addition of 30.00mL OH- Titration of 20.00mL of HC2H3O2(aq) of M=.500M and Ka=1.8x10^-5 with .500M KOH.
Determine the pH before any OH- is added? Also Determine the pH after the addition of 30.00mL OH- Titration of 20.00mL of HC2H3O2(aq) of M=.500M and Ka=1.8x10^-5 with .500M KOH.
Determine the pH before any OH- is added? Also Determine the pH after the addition of 30.00mL OH-
Determine the pH before any OH- is added? Also Determine the pH after the addition of 30.00mL OH-

Explanation / Answer

1) Determine the pH before

The acetic acid will dissociate as

            CH3COOH        -->        CH3COO-      +         H+

Inital             0.5                                    0                  0

change          -x                                x                       x

Equilibrium        0.5-x                           x                    x

So Ka = [H+] [CH3COO-] / [CH3COOH]

1.8 X 10^-5 = x2 / (0.5-x)

As Ka is very small we can ignore x in denominator

1.8 X 10^-5 = x2 / 0.5

0.9 X 10^-5 = x2

x = 3 X 10^-3

[H+] = x = 3 X 10^-3

pH = -log [H+] = 2.52

2) after the addition of 30.00mL OH-

Moles of OH- added = molarity X volume = 0.5 X 30 = 15 millimoles

moles of acid present = molarity X volume = 0.5 X 20 = 10 millimoles

Base will neutralize all of the acid

so moles of base used for neutralization = 10 millmoles

so moles of base left = 5 millimoles

so concentration of base left = millimoles / total volume = 5 / 50 = 0.1 M

we know that

pOH = -log[OH-] = -log [0.1]

pOH = 1

so pH = 14- 1 = 13

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.