Can these question be solved with another method besides the ICE table ? Calcula
ID: 477845 • Letter: C
Question
Can these question be solved with another method besides the ICE table ?
Calculate the pH values and draw the titration curve for the titration of 500 mL of 0.010 M acetic acid (pKa 4.76) with 0.010 MKOH
a) Calculate the pH of the solution after 0 mL of the titrant have been added
b) Calculate the pH of the solution after 250 mL of the titrant have been added.
c) Calculate the pH of the solution after 490 mL of the titrant have been added.
d) Calculate the pH of the solution after 500 mL of the titrant have been added.
e) Calculate the pH of the solution after 510 mL of the titrant have been added.
f) Calculate the pH of the solution after 750 mL of the titrant have been added
Explanation / Answer
millimoles of acetic acid = 500 x 0.01 = 5
pKa = 4.76
a)
pH = 1/2 (pKa - log C)
= 1/2 (4.76 - log 0.01)
= 3.38
pH = 3.38
b)
it is half equivalence point . so here
pH = pKa
pH = 4.76
c)
millimoles of KOH = 490 x 0.01 = 4.9
CH3COOH + KOH -------------> CH3COO- + H2O
5 4.9 0 0
0.1 0 4.9
pH = pKa + log [salt / acid]
= 4.76 + log [4.9 / 0.1]
= 6.45
pH = 6.45
d)
it is half equivalence point . so here salt only remains.
salt concentration = 5 / (500 + 500) = 5 x 10^-3
pH = 7 + 1/2 (pKa + log C)
= 7 + 1/2 (4.76 + log 5 x 10^-3)
= 8.23
pH = 8.23
e)
here base remains = 5.1 - 5 = 0.1
[OH-] = 0.1 / (500 + 510) = 9.9 x 10^-5 M
pOH = -log [OH-] = -log (9.9 x 10^-5)
= 4
pH = 10.0
f)
[OH-] = 2.5 / (500 + 750) = 2 x 10^-3
pOH = 2.70
pH = 11.3
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