6. 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

Now that we have equilibrium concentrations and we can substitute into a K expression, in this case Kf. Which is the correct expression?

a) Kf = [ OH- ][ H+ ]
b) Kf = {[ Cr3+ ][ OH- ]4} / {[ Cr(OH)4- ]}
c) Kf = {[ Cr(OH)4- ]} / {[ Cr3+ ][ OH- ]}
d) Kf = {[ Cr(OH)4- ]} / {[ Cr3+ ][ OH- ]4}