C31 Posted April 6, 2020 Share 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 Link to comment Share on other sites More sharing options...
Solution foramen Posted April 6, 2020 Solution Share 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 Link to comment Share on other sites More sharing options...
C31 Posted April 6, 2020 Author Share Posted April 6, 2020 Merci j’avais pas fais gaffe Quote Link to comment Share on other sites More sharing options...
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