answer the following questions about the titration of 40 ml of .150 M H2SO4 with 1.00 m NaOH. what is the initial pH of the solution before any NaOH is added? what is the pH of the solution at the first equivalence point? what is the pH of the solution at the second equivalence point? what is the relationship between the first equivalence point and the second equivalence point?
Initial pH of 0.150 M of H2SO4
H2SO4 <===> H+ + HSO4 - pKa1 = -3 that mean it is completely ionized. Now we have [H2SO4] = 0 and [H+] = [ HSO4-] = 0.150M
HSO4 - <===> SO4 2- + H+ pKa2 = 1.92, Ka2 = 1.2 × 10-2
[H+] from second dissociation
Ka2 = [ SO4 2-][H+]/ [ HSO4 -]
0.012 = x ( 0.15 + x)/ 0.15-x)
Solving for x
x = 0.0104M = [SO4 2-]
[H+] = 0.150 + 0.0104 = 0.1604 M
pH = -log [H+] = - log 0.1604 = 0.794
Initial pH = 0.794
At first equivalent point that mean
Moles of NaOH = moles of H2SO4
Moles of H2SO4 = 0.15 × 40/ 1000 = 0.006 moles
Moles of NaOH = 0.006 moles
Volume of 1M of NaOH = 6 ml
For neutralization
1 moles of H2SO4 = 2 moles of NaOH
Moles of H2SO4 left = 0.006/2 = 0.003
Molarity of H2SO4 = 0.003 moles in 40 + 6 ml
= 0.003 × 1000/ 46 = 0.065 M
Now H+ after first dissociation = [H+] = [ HSO4 - ] = 0.065M
H+ from Second dissociation
Ka2 = [ H+][SO4 2-]/ [HSO4 -]
0.012 = (0.065 + x)x/ (0.065 - x)
Solving for x
x = 0.009 M = [SO4 2-]
[ H+] = 0.065 + 0.009 = 0.074M
pH = - log 0.074 = 1.13
Second equivalent point
2× Moles of H2SO4 = moles of NaOH
Moles of H2SO4 = 0.006 moles
Moles of NaOH = 0.012 moles
Volume of 1 M NaOH = 12 ml
Total volume =40 + 12 = 52 ml
Now there is no H+ nor OH-. It has only SO4 2- which is behave as weak base
SO4 2- + H2O <===> HSO4 - + OH- kb2 = kw/ ka2
[SO4 2-] = 0.006 moles in 52 ml = 0.006 × 1000/52 = 0.115M
OH- = √(10-14/ 0.012 × 0.115) = 3.00 × 10-7M
pH = 14 + log [OH-] = 7.49
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