Calculate Ka for a weak monoprotic acid?
2010-12-01 7:53 pm
The pH of a 3.90×10-4 M solution of a weak monoprotic acid is 5.31. Calculate Ka for this monoprotic acid to three significant figures.
回答 (2)
2010-12-01 7:57 pm
✔ 最佳答案
[H+]= 10^-5.31 =4.90 x 10^-6 MHA < => H+ + A-
at equilibrium [H+]= [A-] = 4.90 x 10^-6 M
concentration of HA at equilibrium = 3.90 x 10^-4 - 4.90 x 10^-6 =3.85 x 10^-4 M
Ka = [H+][A-]/ [HA] = ( 4.90 x 10^-5) (4.90 x 10^-5) / 3.85 x 10^-4 =6.24 x 10^-6
2010-12-02 4:11 am
pH=-log to the base 10 [H+]
reverse log to get the concentration of [H+]
so 10^-5.31=[H+]
[H+]=4.89...x10^-6
[H+]sq/concentration of the weak acid
so (4.89..x10^-6)sq/(3.90x10^-4)=Ka
Ka=6.15x10^-8
reverse log to get the concentration of [H+]
so 10^-5.31=[H+]
[H+]=4.89...x10^-6
[H+]sq/concentration of the weak acid
so (4.89..x10^-6)sq/(3.90x10^-4)=Ka
Ka=6.15x10^-8
參考: A2 chemistry student.
收錄日期: 2021-04-20 21:56:26
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