Carbonate anions?

Question

I know that carbonate anions can act as either an acid or a base so I

guess these kinds of questions always confused me.

Calculate the pOH of 0.775 M KHCO3, given that the Ka for KHCO3 is 4.7 x 10-11.

I am pretty sure potassium bicarbonate is a base but is there a a

better way for me to determine that - I know K+ is a spectator ion so

you are left with HCO3- but that looks like an oxyacid... so if it is

reacting with water, would HCO3- act as an acid or a base?

Answer 1

Chemgirl? C'est toi, Ernst? (WWI joke about les boches.) If it is you, could you possibly help me get in touch? Am cut off from our fave blog, in a weird way. If you don't recognize my moniker, let me apologize now for messing with your Answers :)

Answer 2

It doesn't really matter if it is an acid or base in order for you to solve the question. Given the Ka, you can easily calculate the pH and from that pOH.

Answer 3

You have 2 possibilities-

HCO3- <--> H^+1 + CO3^-2  

Ka = 4.7x10^-11 = [H+][CO3--] / [HCO3-]

HCO3^-1 + H2O <--> H2CO3 + OH^-1

Kb = Kw/Ka = 2.13x10^-4 = [H2CO3][OH-]/[HCO3-]

Based on Ka and Kb, which equilibrium do you think will predominate?

Hint: The problem gives you a BIG clue...calculate the pOH.

Answer 4

... so if it is reacting with water, would HCO3- act as an acid or a base?  The answer is "YES".  Bicarbonate ion can act as either an acid or a base.  It is amphoteric, which means that it can function as either an or a base.

This occurs in acidic solutions

2HCO3-  +  2H2O <==> 2H3O+ + CO2  + CO3=

This occurs in basic solutions

HCO3-  + H2O <==> H2O + CO2 + OH-

(I know this looks a bit odd, but since there is no H2CO3 in aqueous solution we need to write it as it actually exists, CO2 and H2O.  It would probably be better to write the equilibrium as HCO3-  <==> CO2 + OH-

If you are given Ka for HCO3- then use the equilibrium where it behaves as an acid.

If the equilibrium occurs in basic solution, then use Ka and Kw to compute Kb.

Kw = Ka x Kb

Kb = Kw / Ka

If Ka is very small, then Kb is going to be much greater, relatively, anyway, and the solution of KHCO3 will be basic.

Kb = [CO2] [OH-] / [HCO3-] = Kw/Ka

Kb = (x) (2x)^2 / 0.775 = 2.13 x 10^-4

solve for x

x = 0.0346 = [OH-]

pOH = 1.46

pH = 12.54