8. 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}

What assumption can we make about x to simplify the equation?

a) x >> 0.010M
b) x << 0.010M
c) x << 1.0 x 104 M
d) x4 << 0.010M