Question

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

10-7

Answer 1

Step-by-step 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