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

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,

$K_f=\frac{[AlF_6^{3-}]}{[Al^{3+}][F^-]^6}$

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

$4.0\times 10^{19}=\frac{(x)}{(2-x)\times (1-6x)^6}$

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