Respuesta :
Answer:
molarity=0.21M
Explanation:
First write the balance chemical equation:
[tex]2 S_2O_3^{2-}(aq) + I_3^-(aq) \rightarrow S_4O_6^{2-}(aq) + 3 I^-(aq)[/tex]
Calculate the oxidation number of each element both side:
Reactant sideoxidaion number:
[tex]S=+2[/tex]
[tex]O=-2[/tex]
[tex]I=-\frac{1}{3}[/tex]
product side oxidation number:
[tex]S=+2.5[/tex]
[tex]O=-2[/tex]
[tex]I=-1[/tex]
[tex]n-factor \, of\, S_2O_3^{2-}=2\times(2.5-2)=1[/tex]
[tex]n-factor \, of\, I_3^-=2\times(-1+1/3)=-2/3[/tex] this is change in oxidaion numbe rper atom
oxidaion number per molecule=3×(2\3)=2
n-factor of [tex]I_3^-[/tex] =2
[tex]Normality=molarity \times n-factor[/tex].................................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
[tex]N_1[/tex]=0.5
[tex]V_1[/tex]=25ml
[tex]N_2[/tex]=?
[tex]V_2[/tex]=30ml
[tex]N_1V_1=N_2V_2[/tex]
on putting all the value
[tex]N_2[/tex]=0.42 normality of potassium triiodide
[tex]Normality=molarity \times n-factor[/tex]
molarity=0.42/2=0.21M
molarity=0.21M
