Answer to Question #100442 in General Chemistry for Malathy

Question

A.Standard Sodium Thiosulfate Solution
1.Mass in grams KIO3 in 250mL________g
Aliquot volume__________mL

Sample1 Sample2 Sample3
2.Final buret 26.5mL 34.5mL 29.8mL
reading
3. Initial buret 0 7.2mL 4.9mL
reading
4.Volume of
Na2S2O3 solution 26.5mL 27.3mL 24.9mL
________L _________L ________L

5. Molarity of
Na2S2O3 _________M _________M__________M

Avg.M____________(+_) ___________

6. Standard deviation________________

Answer 1

Technical problems,

V(Na₂S₂O₃)=0,0265L; 0,0273L,0,0249L

0,02623 l

KIO₃ + 6Na₂S₂O₃ + 6H⁺ → 3S₄O₆^2- + I^- + K⁺ + 12Na⁺ + 3H₂O

0,1 normal Na2S2O3=0,1 mole/litr

n(Na₂S₂O₃)=0,00265; 0,00273; 0,00249; moles

0,00262325(+_0,0000025)

0,002623 mole

n(Na₂S₂O₃) = 6 * n(KIO₃ )

n(KIO₃ )=0,002623 /6 =0,0004371moles

V_medium/n_medium =/c_medium

0,02623l0,0004371= 1/ 0,01667

C for aliquote = 0,01667 mole/litr

in V=250ml => 0,0041675 mole

in V=x => 0,0004371

V=x=26,26

m(KIO₃)=mole*MolarWeight=0,8918