Question

A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250 M KSCNC.3.00 mL of 0.050 M HNO3Calculate the initial concentrations of all ions, after mixing, prior to the reaction occurring. The equilibrium concentration of Fe(NCS)2+ was determined using a spectrophotometer to be 3.6 x 10-5 M. Calculate the concentrations of all ions at equilibrium. Calculate the value for the equilibrium constant, K.

Answer 1

Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}$

A. 2.00 mL of 0.00250 M $Fe(NO_3)_3$

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

$n=0.00250 M\times 0.002 L=0.000005 mol$

B. 5.00 mL of 0.00250 M $KSCN$

Moles of KSCN  = n'

Volume of KSCN  = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

$n'=0.00250 M\times 0.005 L=0.0000125 mol$

C. 3.00 mL of 0.050 M $HNO_3$

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

$n=0.050 M\times 0.003 L=0.00015 mol$

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

$[Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M$

Concentration of ferric ions :

$[Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M$

Concentration of nitrate ions from ferric nitrate:

$[NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M$

Concentration of KSCN :

$[KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M$

Concentration of $SCN^-$ ions:

$[SCN^-]=1\times [KSCN]=0.00125 M$

Concentration of potassium ions:

$[K^+]=1\times [KSCN]=0.00125 M$

Concentration of nitric acid :

$[HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M$

Concentration of hydrogen ion :

$[H^+]=1\times [HNO_3]=0.015 M$

Concentration of nitrate ions from nitric acid  :

$[NO_3^{-}]=1\times [HNO_3]=0.0015 M$

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

$Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}$

given concentration of $Fe(NCS)^{2+}$ at equilbrium = $3.6\times 10^{-5} M = 0.000036 M$

initially :

0.0005 M     0.00125 M        0

At equilibrium

(0.0005-0.000036) M   (0.00125-0.000036) M      0.000036 M

0.000464 M     0.001214 M               0.000036 M

The expression of an equilibrium constant will be given as;

$K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}$

$=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91$

The value for the equilibrium constant is 63.91.