white wine 50.00 mL 0.03291 M NaOH 24.57 mL

white wine 50.00 mL 0.03291 M NaOH 24.57 mL

 

 

백포도주 시료 50.00 mL 적정에 0.03291 M NaOH 24.57 mL 소비.

100 mL당 타타르산의 g으로 포도주의 산도를 나타내시오.

 

 

A 50.00 mL sample of a white dinner wine required

24.57 mL of 0.03291 M NaOH to achieve a phenolphthalein end point.

Express the acidity of the wine in grams of tartaric acid

(H2C4H4O6; 150.09 g/mol) per 100 mL.

(Assume that two hydrogens of the acid are titrated.)

 

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적정에 소비된 NaOH의 몰수를 계산하면,

(0.03291 mol/L) (24.57/1000 L) = 0.0008086 mol NaOH

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

Assume that two hydrogens of the acid are titrated

H2C4H4O6(aq) ⇌ 2H^+(aq) + C4H4O6^2-(aq) 이므로,

 

NaOH와 반응한 H2C4H4O6의 몰수

= 0.0008086 mol / 2 = 0.0004043 mol H2C4H4O6

 

 

 

H2C4H4O6의 몰질량 = 150.09 g/mol 이므로,

0.0004043 mol × (150.09 g/mol) = 0.06068 g H2C4H4O6

( 참고 https://ywpop.tistory.com/7738 )

 

---> 50.00 mL 시료에 들어있는 H2C4H4O6의 질량

 

 

 

0.06068 g : 50.00 mL = ? g : 100 mL

 

? = 0.06068 × 100 / 50.00 = 0.1214 g

 

 

 

답: 0.1214 g/100 mL

 

 

 

 

[공식]

NaOH 몰농도 × (NaOH mL/1000) / 2 × 150.09 × 100 / 시료 mL

= 0.03291 × (24.57/1000) / 2 × 150.09 × 100 / 50.00

= 0.1214 g/100 mL

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/19661 ] 0.230 M H2C4H4O6 35.0 mL + 0.150 M NaOH

H2C4H4O6(aq) + 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l)

 

 

 

[키워드] tartaric acid 기준문서