In: Chemistry
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)
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.