Explanation:
A. 2.00 mL of 0.00250 M
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
B. 5.00 mL of 0.00250 M
Moles of KSCN = n'
Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)
Molarity of KSCN = 0.00250 M
C. 3.00 mL of 0.050 M
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
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 :
Concentration of ferric ions :
Concentration of nitrate ions from ferric nitrate:
Concentration of KSCN :
Concentration of ions:
Concentration of potassium ions:
Concentration of nitric acid :
Concentration of hydrogen ion :
Concentration of nitrate ions from nitric acid :
Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M
given concentration of at equilbrium =
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;
The value for the equilibrium constant is 63.91.