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0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 당량점 부피
24.0 mL of 0.105 M acetic acid (HC2H3O2)
will be titrated with 0.130 M sodium hydroxide (NaOH).
The Ka of acetic acid is 1.77×10^(-5).
b) the volume of added base required to reach the equivalence point
Consider the titration of a 24.0 mL sample of 0.105 M CH3COOH
with 0.130 M NaOH. What is the volume of added base
required to reach the equivalence point?
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▶ 참고: 약산-강염기 적정
[ https://ywpop.tistory.com/2736 ]
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M1V1 = M2V2
산의 mol수 = 염기의 mol수
( 참고 https://ywpop.tistory.com/4689 )
(0.105 M) (24.0 mL) = (0.130 M) (? mL)
? = (0.105) (24.0) / (0.130)
= 19.38 mL
[키워드] CH3COOH NaOH 적정 기준문서, 약산 강염기 적정 기준문서
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