Respuesta :
Answer:
The concentrations are :
[tex][HAsc^-]=0.000702 M[/tex]
[tex][Asc^{2-}]=5.92\times 10^{-8} M[/tex]
The pH of the solution is 3.15.
Explanation:
[tex]H_2Asc\rightleftharpoons HAs^-+H^+[/tex] [tex]K_{a1}=1.0\times 10^{-5}[/tex]
Initial
c 0 0
Equilibrium
c-x x x
[tex]K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}[/tex]
[tex]1.0\times 10^{-5}=\frac{x\times x}{(c-x)}[/tex]
[tex]1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}[/tex]
Solving for x:
x = 0.000702 M
[tex][HAsc^-]=0.000702 M[/tex]
[tex]HAsc^-\rightleftharpoons As^{2-}+H^+[/tex] [tex]K_{a2}=5\times 10^{-12}[/tex]
Initially
x 0 0
At equilibrium ;
(x - y) y y
[tex]K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}[/tex]
[tex]5\times 10^{-12}=\frac{y\times y}{(x-y)}[/tex]
[tex]5\times 10^{-12}=\frac{y^2}{(x-y)}[/tex]
Putting value of x = 0.000702 M
[tex]5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}[/tex]
[tex]y=5.92\times 10^{-8} M[/tex]
[tex][Asc^{2-}]=5.92\times 10^{-8} M[/tex]
Total concentration of [tex][H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M[/tex]
The pH of the solution :
[tex]pH=-\log[H^+][/tex]
[tex]pH=-\log[7.0206\times 10^{-4} M}=3.15[/tex]
