Question

What is the molarity of a potassium triiodide solution, KI3(aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.500 M thiosulfate solution, K2S2O3(aq)? The chemical equation for the reaction is 2 S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3 I-(aq)?

a. 0.167 m
b. 0.333 m
c. 0.120 m
d. 0.0833 m

Answer 1

molarity=0.21M

Step-by-step explanation:

First write the balance chemical equation:

`2 S_2O_3^(2-)(aq) + I_3^-(aq) \rightarrow S_4O_6^(2-)(aq) + 3 I^-(aq)`

Calculate the oxidation number of each element both side:

Reactant sideoxidaion number:

`S=+2`

`O=-2`

`I=-(1)/(3)`

product side oxidation number:

`S=+2.5`

`O=-2`

`I=-1`

`n-factor \, of\, S_2O_3^(2-)=2*(2.5-2)=1`

`n-factor \, of\, I_3^-=2*(-1+1/3)=-2/3` this is change in oxidaion numbe rper atom

oxidaion number per molecule=3×(2\3)=2

n-factor of
`I_3^-` =2

`Normality=molarity * n-factor`.................................1

For thiosulfae solution:

Normality of thiosulfate solution=0.5

lets N1 and N2 are the normality of thiosulfate solution and potassium triiodide solution respectively

lets V1 and V2 are the volume of thiosulfate solution and potassium triiodide solution respectively

`N_1`=0.5

`V_1`=25ml

`N_2`=?

`V_2`=30ml

`N_1V_1=N_2V_2`

on putting all the value

`N_2`=0.42 normality of potassium triiodide

`Normality=molarity * n-factor`

molarity=0.42/2=0.21M

molarity=0.21M