Question

1. Given that the Ka1 and Ka2 for carbonic acid are 4.45*10^-7 and 4.69*10^-11, respectively, please determine the alpha value for the HCO₃^- at pH 4.0.

2. Given that the formation constant for Ca-EDTA complex is 5.0*10¹⁰, please determine the conditional formation constant for the complex at pH 8.0 (The alpha4 value for EDTA at pH 11 is 4.2*10^-3).

Answer 1

Actually , Carbonic acid is a solution of dissolved CO2 in water , which behaves as a very weak acid.

The acid ionizes according to following equilibria,

H2CO3 (aq) <----------> HCO3^- (aq ) + H⁺ (aq)

HCO3^- (aq) <------------> CO3^2-   (aq) + H⁺ (aq)

The equilibrium equations for these reactions are labeled as '1' & '2' ,hence

K_a1   = [H⁺ ] [ HCO3^- ] / [ H2CO3 ]..................given as = 4.45 x 10^-7

K_a2   = [ H⁺ ] [ CO3^2- ] / [HCO3^- ] ...............given as = 4.69 x 10^-11

The above two equations when solved gives fraction of carbonates in the following form* representing 'alpha ' as shown in the relation*

Given pH = 4.0

so, 4.0 = -log [ H⁺ ]

& therefore [ H⁺ ] = antilog of ( -4.0 )

.............................= 1.0 x 10^-4 moles / L

Now ,Use the relation*,

_HCO3^-   = [H⁺ ] K_a1 / ( [ H⁺ ]²   + [ H⁺ ] K_a1  + K_a1 K_a2 )

Substituting the given values for K_a1  , K_a2  & [ H⁺ ] = 1.0 x 10^-4 in the above expression & solving for _HCO3^- we get,

_HCO3^- = ( 1 x 10^-4 x 4.45 x 10^-7 ) / { (1 x 10^-4 )² + ( 1 x 10^-4 ) ( 4.45 x 10^-7 ) + ( 4.45 x 10^-7 x 4.69 x 10^-11 ) }

.............. = 4.43 x 10^-3

^{======================================================================================Glad to help. Please post the other question ( ie. Q . 2 ) separately as fresh question.}