0.1180 M KOH 25.98 mL CH3COOH 52.50 mL

0.1180 M KOH 25.98 mL CH3COOH 52.50 mL

 

 

0.1180 M KOH 표준 용액 25.98 mL가

CH3COOH 용액 52.50 mL와 반응한다면

산 용액의 몰농도는 얼마인가?

 

 

If 25.98 mL of a standard 0.1180 M KOH solution

reacts with 52.50 mL of CH3COOH solution,

what is the molarity of the acid solution?

 

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CH3COOH + KOH → CH3COOK + H2O

 

 

 

MV = M’V’

H^+ 이온의 몰수 = OH^- 이온의 몰수

( 참고 https://ywpop.tistory.com/4689 )

 

 

 

(? M) (52.50 mL) = (0.1180 M) (25.98 mL)

? = (0.1180) (25.98) / (52.50) = 0.05839 M

 

 

 

답: 0.05839 M