10,00 = 10.31 + ( log [CO3(-)] / log [HCO3(-)]

Not bad, although you added a log into the equation.

Now it should be a matter of simple concentration calculations. [HCO_{3}^{-}] is given (not directly, but you shouldn't haver any problems calculating it).

H2CO3 + H2O -----> H3O(+) + HCO3(-) Keq1º = 4,45x10^-7

HCO3(-) + H2O -----> H3O(+) + CO3(2-) Keq2º = 4,84x10^-11

10,00 = 10.31 + ( log [CO3(-)] / [HCO3(-)]

Log [CO3(2-)] - Log [HCO3(-)] = 0,31

------------------------------------

HCO3 = 1+ 12+48 = 61g

1 mol-----> 61g

x---------> 5g

(12,2 mol)-???

--------------------------------

Log [CO3(2-)] - Log [12,2] = - 0,31

Log [CO3(2-)]= - 0,31 +1,08

Log [CO3(2-)] = 0,77

10^0,77= 5,88mol/L

5,88:10 = 0,58 mol

CO3---> 60g/L

1 mol --------------60

0,58----------------x

R- 34,8g?