Calculate the pH of 0.215 M carbonic acid (H2CO3). (HINT: do NOT worry about the second H+ in this acid!)

Question

How would I solve this?

Answer 1

Carbonic acid is a weak acid.

H2CO3 + H2O---> HCO3 +H30+

Initial H2CO3=0.215 M
Final H2CO3=0.215-x
Initial HCO3=0
Final HCO3= x
Initial H30+=0
Final H30+= x

ka=4.3 x 10-7=[H30+][HCO3]/H2CO3]

4.3 x 10-7=[x][x]/0.215M-x]

5% rule allows for ignoring -x

4.3 x 10-7=[x][x]/0.215M]
sqrt(4.3 x 10-7 *0.215M])=x

x=H3O+

pH=-log[H3O+]