[santibravo]

Hey guys, can anyone help me with this problem?
---> Calculate the needed volume of NaOH to be added to 300 ml of NaHCO3 0,25M so the buffer solution has a pH=10. (H2CO3 : K1 = 4,3x10^-7 e K2 = 4,7x10^-11)
Sorry if there is some grammatical errors, the problem is originally in portuguese.
Calcule o volume de solução de NaOH 0,350 M a ser adicionado a 300 mL de solução de NaHCO3 0,250 M para que se obtenha um tampão de pH = 10. (H2CO3 : K1 = 4,3x10-7 e K2 = 4,7x10-11).
Thanks

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

[santibravo]

Sorry, didn't know.
I did:
NaHCO3 Na⁺ + HCO3^-      ----> K1
HCO3^- + H20 H3O⁺ + CO3^2-   ---->K2

[H3O⁺]=10^-10 
K₂=[H3O⁺]² / [HCO3^-]

I didn't continue, because I can't see the job of NaOH in this whole thing.
Appreciate the help.

[Borek]

Start calculating ratio of concentrations of HCO₃^-/CO₃^2- at pH=10.