7. 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

Using the expression:

Kf = {[ Cr(OH)4- ]} / {[ Cr3+ ][ OH- ]4},
Which equation is the correct substitution of values?
a) 8.0 x 1029 = {x2} / {(0.010-x)}
b) 8.0 x 1029 = {(0.010-x)} / {(x)(x4)}
c) 8.0 x 1029 = {(x)(x)} / {(x)(1.0 x 10-4)4}
d) 8.0 x 1029 = {(0.010-x)} / {(x)(1.0 x 10-4)4}