Chemistry
posted by Edward .
Calculate the pH of 0.215 M carbonic acid (H2CO3). (HINT: do NOT worry about the second H+ in this acid!)
How would I solve this?

Carbonic acid is a weak acid.
H2CO3 + H2O> HCO3 +H30+
Initial H2CO3=0.215 M
Final H2CO3=0.215x
Initial HCO3=0
Final HCO3= x
Initial H30+=0
Final H30+= x
ka=4.3 x 107=[H30+][HCO3]/H2CO3]
4.3 x 107=[x][x]/0.215Mx]
5% rule allows for ignoring x
4.3 x 107=[x][x]/0.215M]
sqrt(4.3 x 107 *0.215M])=x
x=H3O+
pH=log[H3O+]