Answer:
Concentration of product at equilibrium ;
[tex][H^+]=0.0000229 M[/tex]
[tex][CN^-]=0.0000229 M[/tex]
Explanation:
[tex]HCN(aq)\rightleftharpoons H^+(aq) + CN^-(aq)[/tex]
initially
0.85 M 0 0
(0.85-x)M x x
The equilibrium constant of reaction = [tex]K_c= 6.17\times 10^{-10}[/tex]
The expression of an equilibrium cannot can be written as:
[tex]K_c=\frac{[H^+][CN^-]}{[HCN]}[/tex]
[tex]6.17\times 10^{-10}=\frac{x\times x}{(0.85-x)}[/tex]
Solving for x:
x = 0.0000229
Concentration of product at equilibrium ;
[tex][H^+]=0.0000229 M[/tex]
[tex][CN^-]=0.0000229 M[/tex]
The approximation is valid because [tex]K_c[/tex] is very small.
The concentration of the product at Equilibrium is shown on attached image.
Since
0.85 M 0 0
(0.85-x)M x x
Now the value of x should be
[tex]6.17 \times 10^{-10} = \frac{x\times x}{(0.85-x)}[/tex]
x = 0.0000229
So based on this, the above concentration should be determined.
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