Ka and Kb question?

Question

Because Ka for HCO3 is 4.7x10^-11, Kb for the HCO3 is 2.1x10^-4

what makes this statement false?

Answer 1

for the weak acid HCO3-, the ionization equation that you have the value of Ka for would be:

HCO3- + H2O--> H3O+ + CO32-

So, the Kb value that you have is for the carbonate ion acting as a base by the equation:

CO32- + H2O --> HCO3- + OH-

The Kb is NOT for HCO3- acting as a base.

Answer 2

The Kb of conjugate acid & bases can be solved through this equation.

Kb = Kw/Ka

Recall that an acid is a proton donor. Therefore if HCO3 acts an acid, its conjugate base would be (CO3)2-. And if you were to consider (HCO3)- as a base, its conjugate acid is H2CO3. Therefore in order to get the Kb of (HCO3)-, you will need the Ka of its conjugate acid which is H2CO3.

Answer 3

the better the Ka fee for a substance, the greater this is going to be solid. case in point, HF, it is 7.2 x 10 to the -4 would be more desirable damaging to NH4+, it is 5.6 x 10 to the -10. even nevertheless, the decrease the Kb fee for a substance, the greater solid this is. Ammonia's Kb (a million.8 x 10^-5) is a lot weaker than Pydrine, this is a million.7 x 10-9

Answer 4

Ka=1/Kb or Kb=1/Ka

Answer 5

Lancenigo di Villorba (TV), Italy

I AGREE NOT THE PREVIOUS ANSWER.

MY OWN IS :"You wrong the HCO3- 's Basicity's Constant".

DISCUSSION
This question interests HYDROGENCARBONATE ION, e.g. HCO3-.
This ion acts as an AMPHOTERIC ONE, e.g. it take part to ACID/BASE's Reactions sometimes as ACIDIC REACTIVE sometimes as BASIC ONE.
The former case occurs when it is

HCO3-(aq) <---> CO3--(aq) + H+(aq)
5.6E-11 = Ka2 = |CO3--| * |H+| / |HCO3-|

while the second one interests

HCO3--(aq) + H2O(aq) <---> H2CO3-(aq) + OH-(aq)
2.1E-8 = 1.0E-14 / 4.6E-7 = |H2CO3| * |OH-| / |HCO3-|

This isn't the BASICITY's CONSTANT you argued, e.g. you spoken about

CO3--(aq) + H2O(aq) <---> HCO3-(aq) + OH-(aq)
2.1E-4 = 1.0E-14 / 4.6E-11 = |HCO3-| * |OH-| / |CO3--|

so I shown the Equilibrium's Constants.

I hope this could be clear.