Ancien Responsable Matière Soleneuh Posted February 20, 2019 Ancien Responsable Matière Share Posted February 20, 2019 Re ^^ QCM 8 : Soit une solution aqueuse d'un monoacide de molarité m=0,1 présentant un pH=4 : --> . Le pKa de ce monoacide est égal à 7 VRAI j'ai utilisé la formule ph = 1/2 (pka - logm) donc pka = (ph + 1/2logm) / 1/2 pka = (ph + 1/2 log m) *2 pka = (4 + 1/2 log0,1)*2 pka = (4 + 1/2 *1)*2 = 4,5*2 = 9 ??? Où se trouve mon erreur ? Quote Link to comment Share on other sites More sharing options...
Solution Magnum Posted February 20, 2019 Solution Share Posted February 20, 2019 Salut solenfdo! Attention tu as oublié ton 1/2! pH = 1/2 pka - 1/2 logm --> 1/2pka = pH + 1/2 logm --> 1/2pka = 4 + (1/2 x -1) --> 1/2 pka = 4 - 0,5 --> 1/2 pka = 3,5 --> *2 --> pka = 7 Quote Link to comment Share on other sites More sharing options...
Ancien Responsable Matière Soleneuh Posted February 20, 2019 Author Ancien Responsable Matière Share Posted February 20, 2019 Aaaaaah ok merci @Magnum !! (j'avais oublié le - aussi ^^'''') merci beaucoup génial tu gères Quote Link to comment Share on other sites More sharing options...
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