分析化學題目.拜託大家幫我算一下!!請盡量詳細一點...謝謝

1.
C6H5NH3^+(aq) + H2O(l) ⇌ H3O^+(aq) + C6H5NH2(aq)... (Ka)
Start: [C6H5NH3^+]o = 0.2mM =0.0002 M, [H3O^+]o = [C6H5NH3^+]o= 0 M
At eqm: [C6H5NH3^+]o = (0.0002 - y)M, [H3O^+]o = [C6H5NH3^+]o= y M

Ka = [H3O^+] [C6H5NH2] /[C6H5NH3^+]
y^2 / (0.0002 - y) = 2.51 x10^-5
y^2 + (2.51 x 10^-5)y - (5.02 x 10^-9) = 0

To solve the quadratic equation, we get y = 5.94 x 10^-5

pH = -log(5.94 x 10^-5) = 4.23


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2.
Let B be the formula of the base

At eqm :
pH = 9.28
pOH = 14 - 9.28 = 4.72
[OH^-] = 10^-4.72


B(aq) + H2O(l) ⇌ HB^+(aq) + OH^-(aq) ...(Kb)
Start: [B]o = 0.10 M, [HB^+] = [OH^-] = 0 M
At eqm: [B] = (0.10 - 10^-4.72) M, [HB^+] = [OH^-] = 10^-4.72 M

Kb = [HB^+] [OH^-] / [B] = (10^-4.72)^2 / (0.10 - 10^-4.72) = 3.63 x 10^-9


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3.
[H^+] from H2SO4 = (1.0 x 10^-8) x 2 = 2 x 10^-8 M

Consider the self dissociation of water :
H2O(l) ⇌H^+(aq) + OH^-(aq)
At eqm : [H^+] = [y + (2 x 10^-8)] M, [OH^-] = y M

Kw = [H^+] [OH^-]
[y + (2 x 10^-8)] y = 1 x 10^-14

[y + (2 x 10^-8)] y = 1 x 10^-14
y^2 + (2 x 10^-8)y - (1 x 10^-14) = 0

To solve the quadratic equation, we get y = 9.05 x 10^-8
pH = -log[(9.05 x 10^-8) + (2 x 10^-8)] = 6.96