week 1 workshop question [ENDORSED]

[lily_oneal_2B]

calculate the equilibrium constant for the complete deprotonation of carbonic acid given:

H2CO3(aq) = H+(aq) + HCO3- (aq) Ka=4.4*10^-7

HCO3- = H+(aq) +CO3 2- (aq) Ka=4.7*10^-11

How do you solve a problem like this? For some reason I can't find it in my notes anywhere. This is from Matthew's week 1 workshop

[Grace Chang 1E]

Hi!

You would multiply the two Ka's.

First, a complete deprotonation means you want the equation : H2CO3 --> 2H+ + CO3 (2-). To do that, you add the two reactions so the HCO3- in each equation cancels out.

When you add reactions, you are essentially multiplying their equilibrium constants. Thus, to get the final answer, you do (4.4*10^-7)(4.7*10^-11).

I hope that helps!

[Chem_Mod]

Great answer from Grace.

Also see my lecture (section on polyprotic acids) where I do this example using carbonic acid showing each step and showing K_a1 x K_a2 = K_overall.