4. Calculate the concentration of free Cr3+ ion when 0.o10 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

We are concerned with equilibrium concentrations only, so we must complete an I.C.E. grid.

Assuming that initially all available Cr3+ combines with excess OH- to form Cr(OH)4-, the initial concentration of Cr3+ = 0M, and the concentration of Cr(OH)4- is 0.010M. This assumption is reasonable because there will be mostly Cr(OH)4- (the above equilibrium lies far to the right).

Another known quantity is the OH- concentration, which remains buffered at a constant pH of 10.0. These quantities we can insert in the I.C.E. table immediately.

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

Suppose we let [ Cr3+ ]EQUILIBRIUM = x, what is the change in the [ Cr3+ ] and the change in [ Cr(OH)4- ] respectively?

a) +xM, +xM
b) (0.010-x)M, +xM
c) 0.010M, -xM
d) +xM, -xM