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

Step-by-step 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* 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* 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* 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]=(0.000005 mol)/(0.010 L)=0.0005 M`

Concentration of ferric ions :

`[Fe^(3+)]=1* [Fe(NO_3)_3]=0.0005 M`

Concentration of nitrate ions from ferric nitrate:

`[NO_3^(-)]=3* [Fe(NO_3)_3]=0.0015 M`

Concentration of KSCN :

`[KSCN]=(0.0000125 mol)/(0.010 L)=0.00125 M`

Concentration of
`SCN^-` ions:

`[SCN^-]=1* [KSCN]=0.00125 M`

Concentration of potassium ions:

`[K^+]=1* [KSCN]=0.00125 M`

Concentration of nitric acid :

`[HNO_3]=(0.00015 mol)/(0.010 L)=0.015 M`

Concentration of hydrogen ion :

`[H^+]=1* [HNO_3]=0.015 M`

Concentration of nitrate ions from nitric acid :

`[NO_3^(-)]=1* [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* 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=([Fe(NCS)^(2+)])/([Fe^(3+)][SCN^(-)])`

`=(0.000036 M)/(0.000464 M* 0.001214 M)=63.91`

The value for the equilibrium constant is 63.91.