0.103 M Ni^2+(aq) 10.0 mL 0.0703 M DMG(aq) Ni(DMG)2(s)

0.103 M Ni^2+(aq) 10.0 mL 0.0703 M DMG(aq) Ni(DMG)2(s)

 

 

Ni^2+(aq) 이온은 수용액에서 DMG와 반응하여 아래 반응처럼 침전이 형성된다.

0.103 M Ni^2+(aq) 용액 10.0 mL에 있는 Ni^2+(aq)를

전부 침전시키기 위해 필요한 0.0703 M DMG(aq) 용액의 부피?

Ni^2+(aq) + DMG(aq) → Ni(DMG)2(s) (계수 맞추어 있지 않음)

 

 

The amount of nickel(II) present in an aqueous solution

can be determined by precipitating the nickel

with the organic chemical reagent dimethylglyoxime

[CH3C(NOH)C(NOH)CH3, commonly abbreviated as “DMG”].

Ni^2+(aq) + 2DMG(aq) → Ni(DMG)2(s)

How many mL of 0.0703 M DMG solution is required

to precipitate all the nickel(Il)

present in 10.0 mL of 0.103 M nickel(II) sulfate solution?

 

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Ni^2+(aq) + 2DMG(aq) → Ni(DMG)2(s)

 

 

 

aMV = bM’V’

( a, b는 균형 맞춘 반응식에서 상대방 물질의 계수 )

( 참고 https://ywpop.tistory.com/4689 )

 

(2) (0.103 M) (10.0 mL) = (1) (0.0703 M) (? mL)

 

? = [(2) (0.103) (10.0)] / [(1) (0.0703)]

= 29.3 mL