Question

A mixture is prepared by combining 3.0 mL of 0.0050 M NaSCN solution with 4.0 mL of 0.0030 M Fe(NO3)3 solution, and 3.0 mL of 0.10 M HNO3

a). if Keq = 405, determine the equilibrium concentration of FeSCN2+ in the solution. (keep 4 decimal places)

b). if Keq=282, determine the equilibrium concentration of Fe3+ in the solution. (keep 4 decimal places)

c). if Keq = 343, determine the equilibrium concentration of SCN-- in the solution. (keep 4 decimal places)

Answer 1

Total volume 10 mL

Initial values, before reaction:

[Fe3+] = 0.0030 M x 4mL/10mL = 0.0012 M

[SCN-] = 0.0050 M x 3mL/10 mL= 0.0015 M

Fe3+ + SCN- = FeSCN2+
0.0012M 0.0015 M 0 initial

0 0.0003 M 0.0012M after reaction

X 0.0003+X 0.0012 – X at equilibrium

K = (0.0012 – X) / ( X(0.0003+X)= 405

405X²+ 1.1215X-0.0012= 0

Solving the quadratic equation for X

X= 0.0008. (NeglacNeg the negative X value)

Now,

[FeSCN2+] = (0.0012- X) M.

= 0.0012- 0.0008 M

= 0.0004 M

Similarly, if K= 282

We have, 282= (0.0012- X)/(X(0.0003+X).

(0.0012- X)= 0.0846 X+ 282 X²

Solving the equation to get X= 0.0008.

Therefore, [Fe3+]= X= 0.0008 M.

Also, if K= 343,

We have, 343= (0.0012- X)/(X(0.0003+ X))

X= 0.00085.

Therefore, [SCN-]= 0.0003+X= 0.0003+ 0.00085

= 0.00115 M.