[CO3 2-] = Ka2=5.6×10^-11 mol/L.分析如下:
H2CO3为二元弱酸,其饱和溶液计算可得:
(1)H2CO3 = H+ + HCO3 - Ka1=4.7×10^-7
(2)HCO3 - = H+ + CO3 2- Ka2=5.6×10^-11
由(2)式:[H+][CO3 2-] / [HCO3-] = Ka2
其中[H+] ≈ [HCO3-]
[CO3 2-] = Ka2=5.6×10^-11
[CO3 2-] = Ka2=5.6×10^-11 mol/L.分析如下:
H2CO3为二元弱酸,其饱和溶液计算可得:
(1)H2CO3 = H+ + HCO3 - Ka1=4.7×10^-7
(2)HCO3 - = H+ + CO3 2- Ka2=5.6×10^-11
由(2)式:[H+][CO3 2-] / [HCO3-] = Ka2
其中[H+] ≈ [HCO3-]
[CO3 2-] = Ka2=5.6×10^-11