Fe^2+ 11.00 mL 0.118 M KBrO3 37.50 mL

Fe^2+ 11.00 mL 0.118 M KBrO3 37.50 mL

 

 

Fe^2+ 용액 11.00 mL와 반응하는데

0.118 M KBrO3 용액 37.50 mL가 필요하다면

수용액에서 Fe^2+ 이온의 몰농도는 얼마인가?

알짜 이온 반응식

6Fe^2+(aq) + BrO3^-(aq) + 6H^+(aq)

→ 6Fe^3+(aq) + Br^-(aq) + 3H2O(l)

 

 

What is the molar concentration of Fe^2+ ion in an aqueous solution

if 37.50 mL of 0.118 M KBrO3 is required

for complete reaction with 11.00 mL of the Fe^2+ solution?

The net ionic equation is:

6Fe^2+(aq) + BrO3^-(aq) + 6H^+(aq)

→ 6Fe^3+(aq) + Br^-(aq) + 3H2O(l)

 

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Fe^2+ : BrO3^- = 6 : 1 계수비

 

 

 

aMV = bM’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(1) (? M) (11.00 mL) = (6) (0.118 M) (37.50 mL)

? = [(6) (0.118) (37.50)] / [(1) (11.00)] = 2.4136 M

 

 

 

답: 2.41 M

 

 

 

 

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