Determine the pH of a 0.530 M solution of carbonic acid that has an acid dissociation constant of 4.4 x
10-7

Explanation:

H2CO3  =  H+   +  HCO3-   H2CO3  =  0.530 M  Ka =4.4 X10^-7

Ka =[H+][A-]/[HA]        [H+] = [A-]

0.530 M x 4.4 X10^-7=  [H+}^2

2.32 X10^-7  =  [H+}^2

23.2 X10^-8 = [H+}^2

4.83 X 10^-4 = [H+]

pH = - log 4.83 X 10^-4

pH = -(.68-4)

pH =-(-3.32)

pH= 3.32

answerambulatory;