0.0400 M Hg2(NO3)2 40.0 mL 0.100 M KI

0.0400 M Hg2(NO3)2 40.0 mL 0.100 M KI

 

 

Hg2^2+ + 2I^- → Hg2I2(s) 일 때,

0.0400 M Hg2(NO3)2 40.0 mL와 반응하는데 필요한

0.100 M KI는 몇 mL인가?

 

 

How many milliliters of 0.100 M KI are needed

to react with 40.0 mL of 0.0400 M Hg2(NO3)2

if the reaction is Hg2^2+ + 2I^- → Hg2I2(s)?

 

---------------------------------------------------

 

Hg2^2+(aq) + 2I^-(aq) → Hg2I2(s)

 

Hg2(NO3)2 + 2KI → Hg2I2 + 2KNO3

Hg2(NO3)2 : KI = 1 : 2 계수비(= 몰수비)

 

 

 

aMV = bM’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(2) (0.0400 M) (40.0 mL) = (1) (0.100 M) (? mL)

 

? = [(2) (0.0400) (40.0)] / [(1) (0.100)] = 32.0 mL

 

 

 

답: 32.0 mL

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/12034 ] 0.05 M Hg2(NO3)2 용액 20.0 mL를 Hg2I2(s)로 침전시키기 위해 0.1 M KI로 적정