Question

Test Tube	mL of 0.005M I3-	mL of 0.20M KI	Final Concentration of I3-
1	8.00	0.00
2	6.40	1.60
3	4.80	3.20
4	3.20	4.80
5	2.40	5.60
6	1.60	6.40
7	0.80	7.20
8	0	8.00

What are the final concentrations for I3-?

Answer 1

test tube 1

[KI] or concentration of KI >>>[I3-] ,so its concentration is in excess and remains constant throughout .

KI---->K⁺ +I^-

I3^-<---->I2+I^-

I3- will dissociate negligibly ,due to [I-] high ,common ion effect.The equilibrium shifts in reverse direction

Final conc of I3-=mol of I3- added/total volume of solution=molarity*volume /total vol of solution=0.005 mol/L*0.008L/0.008L=0.005M

test tube 2)

Final conc of I3-=mol of I3- added/total volume of solution=molarity*volume /total vol of solution=0.005 mol/L*0.0064L/(0.0064L+0.0016L)=0.000032M/0.008L=0.004M

test tube 3)

final conc of I3-=mol of I3- added/total volume of solution=molarity*volume /total vol of solution=0.005 mol/L*0.0048L/(0.0048L+0.0032L)=0.003M

test tube 4)

final conc of I3-=mol of I3- added/total volume of solution=molarity*volume /total vol of solution=0.005 mol/L*0.0032L/(0.0048L+0.0032L)=0.002M