Respuesta :
Explanation:
[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}[/tex]
A. 2.00 mL of 0.00250 M [tex]Fe(NO_3)_3[/tex]
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
[tex]n=0.00250 M\times 0.002 L=0.000005 mol[/tex]
B. 5.00 mL of 0.00250 M [tex]KSCN[/tex]
Moles of KSCN = n'
Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)
Molarity of KSCN = 0.00250 M
[tex]n'=0.00250 M\times 0.005 L=0.0000125 mol[/tex]
C. 3.00 mL of 0.050 M [tex]HNO_3[/tex]
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
[tex]n=0.050 M\times 0.003 L=0.00015 mol[/tex]
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 :
[tex][Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M[/tex]
Concentration of ferric ions :
[tex][Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M[/tex]
Concentration of nitrate ions from ferric nitrate:
[tex][NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M[/tex]
Concentration of KSCN :
[tex][KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M[/tex]
Concentration of [tex]SCN^-[/tex] ions:
[tex][SCN^-]=1\times [KSCN]=0.00125 M[/tex]
Concentration of potassium ions:
[tex][K^+]=1\times [KSCN]=0.00125 M[/tex]
Concentration of nitric acid :
[tex][HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M[/tex]
Concentration of hydrogen ion :
[tex][H^+]=1\times [HNO_3]=0.015 M[/tex]
Concentration of nitrate ions from nitric acid :
[tex][NO_3^{-}]=1\times [HNO_3]=0.0015 M[/tex]
Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M
[tex]Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}[/tex]
given concentration of [tex] Fe(NCS)^{2+}[/tex] at equilbrium = [tex]3.6\times 10^{-5} M = 0.000036 M[/tex]
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;
[tex]K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}[/tex]
[tex]=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91 [/tex]
The value for the equilibrium constant is 63.91.
