황산 시료 중화에 필요한 0.225 M KOH 용액의 부피

0.225 M H2SO4 45 mL

0.125 M H2SO4 185 mL

0.100 M H2SO4 75 mL

Determine the volume of 0.225 M KOH solution

required to neutralize each sample of sulfuric acid.

The neutralization reaction is:

H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)

a) 45 mL of 0.225 M H2SO4

b) 185 mL of 0.125 M H2SO4

c) 75 mL of 0.100 M H2SO4

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aMV = bM’V’

H^+의 mol수 = OH^-의 mol수

( 참고 https://ywpop.tistory.com/4689 )

a) 0.225 M H2SO4 45 mL

(2) (0.225 M) (45 mL) = (1) (0.225 M) (? mL)

? = [(2) (0.225) (45)] / [(1) (0.225)]

= 90 mL

b) 0.125 M H2SO4 185 mL

? = [(2) (0.125) (185)] / [(1) (0.225)]

= 205.6 mL

c) 0.100 M H2SO4 75 mL

? = [(2) (0.100) (75)] / [(1) (0.225)]

= 66.67 mL

[ 관련 예제 https://ywpop.tistory.com/3973 ] 0.427 M KOH 용액 60.0 mL를 중화하려면 1.28 M H2SO4 용액 몇 mL가 필요한가?

[키워드] 0.225 M KOH 0.225 M H2SO4 45 mL 0.125 M H2SO4 185 mL 0.100 M H2SO4 75 mL, H2SO4 + KOH 중화 기준