C31 Posted April 6, 2020 Posted April 6, 2020 (edited) Bonjour Dans le qcm2, je ne comprend pas d’où sort le 4 dans le calcul de pH de la solution 3 (item D) -log m = -log (0,1) = 1 non ? donc pour moi le calcul est pH= 1/2 (3+1) = 2 si qqn pouvait m’expliquer. merciii Edited April 6, 2020 by C31 Quote
Solution foramen Posted April 6, 2020 Solution Posted April 6, 2020 La concentration est en mmol donc faut mettre 10^-3 ce qui fait pH=1/2*3 - 1/2*log(10^-4) = 1,5 + 2 =3,5 Quote
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