Hello!

The last problem with pH. Solutions of NaH

_{2}PO

_{4} and Na

_{2}HPO

_{4} (both are 0,1 mol/dm

^{3}) are mixed together in ratio 1 : 1. Values given:

K

_{a}(H

_{3}PO

_{4}) = 6*10

^{-3}K

_{a}(H

_{2}PO

_{4}^{-} = 6*10

^{-8}K

_{a}(HPO

_{4}^{2-}) = 5*10

^{-13}I tried to solve it in the following way:

Dissociation:

NaH

_{2}PO

_{4} Na

^{+} + H

_{2}PO

_{4}^{-}Na

_{2}HPO

_{4} 2Na

^{+} + HPO

_{4}^{2-}And then, according to Bronsted-Lowry's theory:

H

_{2}PO

_{4}^{-} + H

_{2}O

HPO

_{4}^{2-} + H

_{3}O

^{+}And now I think that I can calculate the concentration of H

_{2}PO

_{4}^{-} and HPO

_{4}^{2-} after dissociation using ICE table, or maybe dissociation constants (though I don't know how to use them in this case), and than use it to calculate the concentraltion of H

_{3}O

^{+} from K

_{a} (H

_{2}PO

_{4}^{-}).

I don't know if this is correct so please check me.