0.1809M 的 NaHCO3 的 pH 值是多少？

Equilibrium constants:
H2CO3(aq) ≒ HCO3-(aq) + H+(aq) KK1 = 4.3 x 10-7­ M
HCO3-(aq) ≒ CO32-(aq) + H+(aq) KK2 = 5.6 x 10-11 M

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HCO3-(aq) acts as both an acid and a base.
In the solution, the following two equilibria are established.
HCO3-(aq) ≒ CO32-(aq) + H+(aq) KEqm constant = K2
HCO3-(aq) + H+(aq) ≒ H2CO3(aq) KEqm constant = 1/K1

Adding up the above two equations, a combined equilibrium is obtained:
2HCO3-(aq) ≒ CO32-(aq) + H2CO3(aq) KEqm constant = K2/K1
At eqm: [CO32-] = [H2CO3] = y M
At eqm: [HCO3-] = (0.1809 - y) M
y2/(0.1809 - y)2 = K2/K1 = (5.6 x 10-11)/(4.3 x 10-7)
y/(0.1809 - y) = 0.01141
y = 2.04 x 10-3

Consider the following equilibrium:
HCO3-(aq) ≒ CO32-(aq) + H+(aq) KEqm constant = K2
At eqm: [CO32-] = y M = 2.04 x 10-3 M
At eqm: [HCO3-] = (0.1809 - y) M = 0.1789 M
K2 = [CO32-][H+]/[HCO3-]
[H+] = K2[HCO3-]/[CO32-]
[H+] = (5.6 x 10-11) x 0.1789 / (2.04 x 10-3)
[H+] = 4.91 x 10-9 M

pH = -log[H+]
pH = -log(4.91 x 10-9)
pH = 8.31