아스피린 역적정. aspirin 시료 3 g에 1 M NaOH (f=1.020) 50 mL

아스피린 역적정. aspirin 시료 3 g에 1 M NaOH (f=1.020) 50 mL

 

 

aspirin (MW=180) 시료 3 g에 1 M NaOH (f=1.020) 50 mL를 넣고 반응시킨 후 0.5 M H2SO4 (f=1.010)로 역적정하여 20 mL가 소비되었다면 aspirin의 함량은 몇 %인가?

 

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0.5 M H2SO4 (f=1.010) 20 mL에 들어있는 H2SO4의 몰수를 계산하면,

(0.5 mol/L) × 0.020 L × 1.010 = 0.0101 mol H2SO4

 

 

1 M NaOH (f=1.020) 50 mL에 들어있는 NaOH의 몰수를 계산하면,

(1 mol/L) × 0.050 L × 1.020 = 0.051 mol NaOH

 

 

H2SO4 + 2NaOH → Na2SO4 + 2H2O

( 참고 https://ywpop.tistory.com/8047 )

 

H2SO4 : NaOH = 1 : 2 계수비(= 몰수비) 이므로,

0.0101 mol H2SO4와 반응한 NaOH의 몰수를 계산하면,

H2SO4 : NaOH = 1 : 2 = 0.0101 mol : ? mol

? = 2 × 0.0101 = 0.0202 mol NaOH

 

 

aspirin과 반응한 NaOH의 몰수

= 0.051 mol - 0.0202 mol = 0.0308 mol NaOH

 

 

aspirin : NaOH = 1 : 2 반응 (saponification) 하므로,

0.0308 mol NaOH와 반응한 aspirin의 몰수를 계산하면,

aspirin : NaOH = 1 : 2 = ? mol : 0.0308 mol

? = 0.0308 / 2 = 0.0154 mol aspirin

 

 

0.0154 mol × (180 g/mol) = 2.772 g aspirin

 

 

(2.772 / 3) × 100 = 92.4%

 

 

답: 92.4%

 

 

 

 

 

 

[키워드] 아스피린 기준문서, 역적정 기준문서, aspirin saponification

 

 

 

[참고] 아스피린의 산-염기 반응과 비누화 반응

Aspirin is just the common name for acetylsalicylic acid. It is a benzoic acid, substituted in the ortho position with an acylated alcohol (actually a phenol) function.

 

Thus, there are two reactions that can occur between aspirin and NaOH: the faster will be deprotonation of the acid function, so you'll get sodium acetylsalicylate. This is actually what is present in many aspirin formulations.

( 관련 예제 https://ywpop.tistory.com/13926 )

 

The other is saponification of the acetyl function. This function is an ester. In the case of aspirin, it is the salicylate ester of acetic acid. Salicylate is the common name of 2-hydroxy-benzoic acid. So if you saponify the acetyl, you will get sodium acetate and sodium salicylate.

 

The first reaction is complete and fast at room temperature. The second is slower, and requires prolonged heating, and higher NaOH concentration.