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Q.

Given H2SO4strongacid + H2O  H2O+ + HSO4

HSO4weak acid+ H2O  H3O+ + SO42-, Ka = 1.1 × 10-2

Thus, at equilibrium, [H3O+], [HSO4] and [SO42-] in 0.50 M H2SO4 respectively, are

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a

[H3O+]                       [HSO4]                  [SO42-]

0.51M                          0.49M                       0.01M

b

0.49M                0.49M             0.01M

c

0.50M               0.50M              0.50M

d

0.51M               0.51M                0.51M

answer is A.

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Detailed Solution

[H2SO4] = 0.50M      [H3O+] = 0.5 M    [HSO4] = 0.50M

HSO4 H2O+  + SO42-

0.5                       0.5M                0

Question Image

[H3O+] = (0.5 + x) M[SO42-] = xm

H2SO4 = (0.5  x)M    Ka = [H3O+][SO42-][HSO4-] x = 0.01

[H3O+] = 0.51 M [SO42-] = 0.01M [HSO4] = 0.49M

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Given H2SO4strongacid + H2O → H2O+ + HSO4−HSO4− weak acid+ H2O ⇌ H3O+ + SO42-, Ka = 1.1 × 10-2Thus, at equilibrium, [H3O+], [HSO4−] and [SO42-] in 0.50 M H2SO4 respectively, are