Chemical Equilibrium - Buffer

a)
At equilibrium:
pH = pKa + log([A^-]/[HA])
6 = 6 + log([A^-]/[HA])
Hence, [A^-] = [HA]

Let V cm³ be the volume of 1 M HA used.
Then volume of 1M NaOH used = (100 - V) cm³

As Ka is small and the common ion effect, the ionization of HA can beneglected.

HA(aq) + OH^-(aq) → A^-(aq) + H2O(l)
OH^- is completed reacted.
No. of moles of OH^- reacted = 1 x (100 - V)/1000 = (0.1 - 0.001V) mol
No. of moles of A^- formed = (0.1 - 0.001V) mol
No. of moles of HA left = 1 x (V/1000) - 1 x [(100 - V)/1000] = (0.002V - 0.1)mol

At equilibrium:
[A^-] = [HA]
(0.1 - 0.001V) / (100/1000) = (0.002V - 0.1) / (100/1000)
0.1 - 0.001V = 0.002V - 0.1
0.003V = 0.2
V = 66.7

Volume of HA used = 66.7 cm³
Volume of NaOH used = 100 - 66.7 = 33.3 cm³


b)
At equilibrium:
pH = pKa + log([A^-]/[HA])
6 = 6 + log([A^-]/[HA])
Hence, [A^-] = [HA]

Let V cm³ be the volume of 1 M NaA used.
Then volume of 1 M HCl used = (100 - V) cm³

As Ka is small and the common ion effect, the ionization of HA can beneglected.

A^-(aq) + H^+(aq) → HA(aq)
H^+ is completed reacted.
No. of moles of H^+ reacted = 1 x (100 - V)/1000 = (0.1 - 0.001V) mol
No. of moles of HA formed = (0.1 - 0.001V) mol
No. of moles of A^- left = 1 x (V/1000) - 1 x [(100 - V)/1000] = (0.002V - 0.1)mol

At equilibrium:
[A^-] = [HA]
(0.002V - 0.1) / (100/1000) = (0.1 - 0.001V) / (100/1000)
0.002V - 0.1 = 0.1 - 0.001V
0.003V = 0.2
V = 66.7

Volume of NaA used = 66.7 cm³
Volume of HCl used = 100 - 66.7 = 33.3 cm³

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