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Biophysique: P16/17


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salut

 

B. pour le diacide fort c'est pH = -log(2 x 10-2) = - log 2 + - log (10-2) = -0,2 + 2 = 1,7

 

C. pour l'acide faible 1/2(pKa - log m) 

pKa = -log(Ka)= - log(10-5)= 5

pH = 1/2(5 - log (10-2) = 1/2 (5 - (-2)) = 1/2 ( 5 + 2 ) = 1/2 (7) = 3,5

 

D. on fait pareil et on prend pour pKa la constante de premiÚre acidité donc pKa = 7

pH = 1/2(7 - log (10-2) = 1/2 (5 - (-2)) = 1/2 ( 7 + 2 ) = 1/2 (9) = 4,5

 

j'espùre que ça t'aidera 🐾 

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