0.1017 g KBrO3 39.75 mL Na2S2O3

0.1017 g KBrO3 39.75 mL Na2S2O3

0.1017 g KBrO3 HCl KI 39.75 mL Na2S2O3

 

 

A 0.1017 g sample of KBrO3 was dissolved in dilute HCl

and treated with an unmeasured excess of KI.

The liberated iodine required 39.75 mL of a sodium thiosulfate solution.

Calculate the molar concentration of the Na2S2O3.

Calculate the molarity of the Na2S2O3.

 

BrO3^- + 6I^- + 6H^+ → Br^- + 3I2 + 3H2O

 

I2 + 2S2O3^2- → 2I^- + S4O6^2-

 

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KBrO3의 몰질량 = 167.00 g/mol 이므로,

0.1017 g / (167.00 g/mol) = 0.00060898 mol KBrO3

( 참고 https://ywpop.tistory.com/7738 )

 

= 0.00060898 mol BrO3^-

 

 

 

BrO3^- + 6I^- + 6H^+ → Br^- + 3I2 + 3H2O

 

BrO3^- : I2 = 1 : 3 계수비(= 몰수비) 이므로,

BrO3^- : I2 = 1 : 3 = 0.00060898 mol : ? mol

 

? = 3 × 0.00060898 = 0.0018269 mol I2

 

 

 

I2 + 2S2O3^2- → 2I^- + S4O6^2-

( 참고 https://ywpop.tistory.com/14662 )

 

I2 : S2O3^2- = 1 : 2 = 0.0018269 mol : ? mol

 

? = 2 × 0.0018269 = 0.0036538 mol S2O3^2-

= 0.0036538 mol Na2S2O3

 

 

 

몰농도 = 용질 mol수 / 용액 L수

= 0.0036538 mol / (39.75/1000 L)

= 0.09192 mol/L

 

 

 

답: 0.09192 M