9. Calculate the concentration of free Cr3+ ion when 0.010 mol of Cr(NO3)3 is dissolved in a litre of solution buffered at pH = 10.0, and the Kf for Cr(OH)4- = 8.0 x 1029.

Equilibrium equation:

Cr3+ (aq) + 4 OH- (aq) Cr(OH)4- (aq)

[ OH- ] = {1.0 x 10-14} / {10-pH} = 1.0 x 10-4 M

I.C.E. grid:

Cr3+ (aq) 4 OH- (aq) Cr(OH)4- (aq)
INITIAL
0M
1.0 x 10-4 M
0.010M
CHANGE
+xM
buffered (no change)
-xM
EQUILIBRIUM
xM
1.0 x 10-4 M
(0.010-x)M

Kf = {[ Cr(OH)4- ]} / {[ Cr3+ ][ OH- ]4},

8.0 x 1029 = {(0.010-x)} / {(x)(1.0 x 10-4)4}

x = 1.3 x 10-16 M
What does this x represent?
a) the final [ Cr3+ ] (aq) = [ Cr3+ ]EQUILIBRIUM
b) the [ Cr3+ ]EQUILIBRIUM not equal to the final [ Cr3+ ] (aq)
c) the final [ Cr(NO3)3 ]EQUILIBRIUM
d) the final [ Cr(OH)4- ]EQUILIBRIUM