Question

The initial concentration of fluoride ions in an aqueous solution is 2.00 M and the initial concentration of Al3+ ions is 0.15 M. After the solution has reached equilibrium what is the concentration ofAl3+, F- , and AlF6 3- ? Kf for [AlF6] 3- = 4.0 x 1019 .

Answer 1

Answer:
`[Al^(3+)]` = 1.834 M

`[F^-]` = 0.004 M

`[AlF_6^(3-)]` = 0.166 M

Step-by-step explanation:

`Al^(3+)+6F^-\rightleftharpoons AlF_6^(3-)`

Initial concentration of
`Al^(3+)` = 0.15 M

Initial concentration of
`F^-` = 2.0 M

The given balanced equilibrium reaction is,

`Al^(3+)+6F^-\rightleftharpoons AlF_6^(3-)`

Initial conc. 2 M 0.15 M 0

At eqm. conc. (2-x) M (1-6x) M (x) M

The expression for equilibrium constant for this reaction will be,

`K_f=([<strong>AlF_6^(3-)</strong>])/([<strong>Al^(3+)</strong>][F^-]^6)`

Now put all the given values in this expression, we get :

`4.0* 10^(19)=((x))/((2-x)* (1-6x)^6)`

By solving the term 'x', we get :

`x=0.166`

`[Al^(3+)]` = (2-x) = 2-0.166 = 1.834 M

`[F^-]` = (1-6x) = 1-6(0.1660)= 0.004 M

`[AlF_6^(3-)]` = x = 0.166 M