Answer :
Explanation:
The value of first dissociation constant;
[tex]pK_{a1}=-\log[K_{a1}][/tex]
[tex]6.35=-\log[K_{a1}][/tex]
[tex]K_{a1}=4.467\times 10^{-7}[/tex]
The value of seconddissociation constant;
[tex]pK_{a2}=-\log[K_{a2}][/tex]
[tex]10.33=-\log[K_{a2}][/tex]
[tex]K_{a2}=4.467\times 10^{-11}[/tex]
The pH of the water = 7.5
[tex]pH=\log[H^+][/tex]
[tex]7.5=\log[H^+][/tex]
[tex][H^+]=3.162\times 10^{-8} M[/tex]
[tex]H_2CO_3\rightleftharpoons HCO_3^{-}+H^+[/tex]
C 0 0
At equilibrium
(C-x) x x
[tex]HCO_3^{-}\rightleftharpoons CO_3^{2-}+H^+[/tex]
x 0 0
At equilibrium
(x -y) y y
Expression of an second dissociation constant will be given as:
[tex]K_{a2}=\frac{y\times y}{(x-y)}[/tex]
[tex]4.677\times 10^{-11}=\frac{y^2}{(x-y)}[/tex]..[1]
[tex]x+y=[H^+][/tex]
[tex]x+y=3.162\times 10^{-8} [/tex]...[2]
Solving [1] and [2]:
x = [tex]3.045\times 10^{-8} M[/tex]
y = [tex]1.1702\times 10^{-9} M[/tex]
Expression of an first dissociation constant will be given as:
[tex]K_{a1}=\frac{x\times x}{(C-x)}[/tex]
[tex]4.467\times 10^{-7}=\frac{x^2}{(C-x)}[/tex]
[tex]4.467\times 10^{-7}=\frac{(3.045\times 10^{-8} M)^2}{(C-(3.045\times 10^{-8} M))}[/tex]
Solving for C:
[tex] C = 3.253\times 10^{-8} M[/tex]
At equilibrium , concentration of species:
Carbonic acid :
[tex][H_2CO_3]=(C-x)=3.253\times 10^{-8} M-3.045\times 10^{-8} M[/tex]
[tex][H_2CO_3]=2.08\times 10^{-9} M[/tex]
Carbonate ion :
[tex][CO_3^{2-}]=y=1.1702\times 10^{-9} M[/tex]
Bicarbonate :
[tex][HCO_3^{-}]=(x-y)=3.045\times 10^{-8} M-1.1702\times 10^{-9} M=2.928\times 10^{-8} M[/tex]Total carbonates:[TC]
[tex][TC]=[H_2CO_3]+[HCO_3^{-}]+[CO_3^{2-}]=C[/tex]
[tex]= [TC} = 3.253\times 10^{-8} M[/tex]