[Norngpinky]

So my friend and I have been trying to figure out these problems for about a few hours now.

Calculate the concentrations of H+, HCO3‾, and CO32- in a 0.026 M H2CO3 solution.

Ka for H2CO3 = 4.3E-7 and HCO3- = 5.6E-11

So we set up the equation is two steps -- as diprotic acid.
We solved for the concentration of HCO3- to be 4.3E-7 = x^2/H2C03-x using approximation and x=1.057E-4

X equates conc of H+ and HCO3- in step one

Then we used that as the starting conc of HCO3- in step two and use Ka for HCO3- = 5.6E-11 ---> x equates conc of H+ and CO3-2 in step two, which is 7.7E-8

So I got 1.1E-4 for H+ and HCO3- correct, but 7.7E-8 is wrong for CO3-2

So I have no idea what I did wrong there.


Use the data in Table 15.3 to calculate the equilibrium constant for the following reaction.

HCOOH(aq)+ OH -(aq) <--> HCOO-(aq)+ H2O(l)

Ka for HCOOH = 1.7E-4 and Kb for HCOO- = 5.9E-11

I know Kw=Ka*Kb, however I'm not sure how the equilibrium constant comes into play with this.

I tried Ka*Kb which is 1E-14, but that is wrong.

=S

[Borek]

x equates conc of H+ and CO3-2 in step two, which is 7.7E-8

Do I understand correctly that you assumed [CO₃^2-] = [H⁺]? Think again - what about [H⁺] from the first dissociation step?

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified

HCOOH(aq)+ OH -(aq) <--> HCOO-(aq)+ H2O(l)[/b]

Ka for HCOOH = 1.7E-4 and Kb for HCOO- = 5.9E-11

Write formulas for Ka, Kb and Kw - see if you can combine right sides to get reaction quotient for this reaction.