{"id":12238,"date":"2016-01-30T04:20:20","date_gmt":"2016-01-30T03:20:20","guid":{"rendered":"https:\/\/kamerpower.com\/?p=12238"},"modified":"2025-01-29T21:06:02","modified_gmt":"2025-01-29T20:06:02","slug":"corrige-concours-ensp-yaounde-2015-chimie-et-physique","status":"publish","type":"post","link":"https:\/\/kamerpower.com\/fr\/corrige-concours-ensp-yaounde-2015-chimie-et-physique\/","title":{"rendered":"Corrige Concours ENSP Yaounde 2015 Chimie et Physique"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_82_2 counter-hierarchy ez-toc-counter ez-toc-custom ez-toc-container-direction\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Contents<\/p>\n<label for=\"ez-toc-cssicon-toggle-item-69e492b604c35\" class=\"ez-toc-cssicon-toggle-label\"><span class=\"ez-toc-cssicon\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/label><input type=\"checkbox\"  id=\"ez-toc-cssicon-toggle-item-69e492b604c35\"  aria-label=\"Toggle\" \/><nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/kamerpower.com\/fr\/corrige-concours-ensp-yaounde-2015-chimie-et-physique\/#corrige-concours-ensp-yaounde-2015-chimie-et-physique-%e2%80%93-ecole-nationale-polytechnique-de-yaounde\" >Corrige Concours ENSP Yaounde 2015 Chimie et Physique &#8211; Ecole Nationale Polytechnique de Yaound\u00e9<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/kamerpower.com\/fr\/corrige-concours-ensp-yaounde-2015-chimie-et-physique\/#probleme-1\" >Probl\u00e8me\u00a01<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/kamerpower.com\/fr\/corrige-concours-ensp-yaounde-2015-chimie-et-physique\/#solution-au-probleme-1-%e2%80%93-corrige-concours-ensp-yaounde-2015-chimie-et-physique\" >Solution au probl\u00e8me 1 \u00a0&#8211;\u00a0Corrige Concours ENSP Yaounde 2015 Chimie et Physique\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/kamerpower.com\/fr\/corrige-concours-ensp-yaounde-2015-chimie-et-physique\/#kamerpower-probleme-2-%e2%80%93-corrige-concours-ensp-yaounde-2015-chimie-et-physique\" >Kamerpower Probl\u00e8me\u00a02 &#8211;\u00a0Corrige Concours ENSP Yaounde 2015 Chimie et Physique\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/kamerpower.com\/fr\/corrige-concours-ensp-yaounde-2015-chimie-et-physique\/#kamerpower-solution-au-probleme-2-%e2%80%93-corrige-concours-ensp-yaounde-2015-chimie-et-physique\" >Kamerpower Solution au probl\u00e8me 2 \u00a0&#8211;\u00a0Corrige Concours ENSP Yaounde 2015 Chimie et Physique\u00a0<\/a><\/li><\/ul><\/nav><\/div>\n<h3 class=\"post-title single-title entry-title\" style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"corrige-concours-ensp-yaounde-2015-chimie-et-physique-%e2%80%93-ecole-nationale-polytechnique-de-yaounde\"><\/span><span style=\"color: #ff0000;\"><strong>Corrige Concours ENSP Yaounde 2015 Chimie et Physique &#8211; Ecole Nationale Polytechnique de Yaound\u00e9<\/strong><\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<hr \/>\n<h3 class=\"post-title single-title entry-title\" style=\"text-align: center;\"><\/h3>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"probleme-1\"><\/span><strong><span style=\"color: #3366ff;\">Probl\u00e8me\u00a01<\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Corrige Concours ENSP Yaounde<\/strong>: Le paquebot <span style=\"color: #ff0000;\">Narcisse<\/span> aborde la c\u00f4te anglaise par temps calme, dans un \u00e9pais de brouillard, et se dirige droit vers la falaise \u00e0 une vitesse <span style=\"color: #ff0000;\">V<\/span> constante. Il envoie toutes les minutes un coup de sir\u00e8ne bref. kamerpower.com<\/span><\/p>\n<ol>\n<li style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"color: #000000;\">Sachant que la vitesse du son dans l\u2019air est <span style=\"color: #ff0000;\">n=330 m.s<sup>-1<\/sup><\/span>, et que le premier \u00e9cho est entendu au bout d\u2019un temps <span style=\"color: #ff0000;\">T<sub>1<\/sub>=18 s<\/span> apr\u00e8s le premier coup de sir\u00e8ne\u00a0; le deuxi\u00e8me \u00e9cho au bout d\u2019un temps <span style=\"color: #ff0000;\">T<sub>2<\/sub>=15 s<\/span> apr\u00e8s le deuxi\u00e8me coup de sir\u00e8ne, Calculer la vitesse <span style=\"color: #ff0000;\">V<\/span> du paquebot, ainsi que la distance <span style=\"color: #ff0000;\">D<\/span> qui s\u00e9pare la falaise du paquebot au moment o\u00f9 celui-ci envoie le premier coup de sir\u00e8ne.\u00a0<\/span><\/span><span style=\"color: #000000;\">(on appellera <span style=\"color: #ff0000;\">A<sub>1<\/sub><\/span> le point o\u00f9 est \u00e9mis le premier coup de sir\u00e8ne, et <span style=\"color: #ff0000;\">A<sub>2<\/sub><\/span> le point o\u00f9 est \u00e9mis le deuxi\u00e8me coup de sir\u00e8ne, <span style=\"color: #ff0000;\">A<sub>1<\/sub>&#8216;<\/span> le point o\u00f9 est entendu le premier \u00e9cho, <span style=\"color: #ff0000;\">A<sub>2<\/sub>&#8216;<\/span> le point o\u00f9 est entendu le deuxi\u00e8me \u00e9cho.)<\/span><\/li>\n<li style=\"text-align: justify;\"><span style=\"color: #000000;\">Sachant que l\u2019appareil auditif humain est incapable de distinguer deux sons successifs lui parvenant dans un intervalle de temps <span style=\"color: #ff0000;\">dT<\/span> inf\u00e9rieur \u00e0 <span style=\"color: #ff0000;\">0,1 s<\/span>, quelle est la distance minimale <span style=\"color: #ff0000;\">d<\/span> au dessus de laquelle on n\u2019entend plus d\u2019\u00e9cho?<\/span><\/li>\n<\/ol>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"solution-au-probleme-1-%e2%80%93-corrige-concours-ensp-yaounde-2015-chimie-et-physique\"><\/span><strong><span style=\"color: #ff0000;\">Solution au probl\u00e8me 1 \u00a0&#8211;\u00a0<span style=\"color: #000000;\">Corrige Concours ENSP Yaounde 2015 Chimie et Physique\u00a0<\/span><\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"color: #3366ff;\"><strong>1) <\/strong><\/span>Pour V et D\u00a0on entend l\u2019\u00e9cho que lorsque le son se propage dans l\u2019air, se r\u00e9fl\u00e9chit sur la falaise et rebrousse chemin jusqu\u2019au paquebot.\u00a0L\u2019origine des temps est pris en A, il est clair que\u00a0A_{{1}}A_{{1}} ^{&#8216;} = T_{{1}}V \u00a0 \u00a0et \u00a0 A_{{2}}A_{{2}} ^{&#8216;} = T_{{2}}V\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Le son \u00e9mis en A parcourt la distance\u00a02D &#8211; T_{{1}}V \u00a0avec une vitesse v, donc\u00a0T_{1}v= 2D &#8211; T_{{1}}V &#8212;&#8212;(1)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Le son du deuxi\u00e8me coup de sir\u00e8ne a parcouru\u00a0 la distance\u00a0OA_{{2}} ^{&#8216;} +OA_{2}\u00a0\u00a0en\u00a0T_{{2}}\u00a0secondes.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"color: #ff0000;\">Donc:\u00a0<span style=\"color: #000000;\">OA_{{2}} ^{&#8216;} +OA_{2}=T_{{2}}v<\/span><\/span><\/span><\/p>\n<p>OA_{{2}} =D-A_{1}A_{2}=D-\\Delta t.V ; &#8212;\\Delta t=60 s = 1min<\/p>\n<p>A_{{2}} ^{&#8216;} =D- A_{{2}}&#8217;A_{{2}}-A_{{1}}A_{{2}}=D-T_{{2}}V-\\Delta tV=D-(T_{2}+\\Delta t)V<\/p>\n<p>2D-(T_{2}+2\\Delta t)V=T_{{2}}v &#8212;&#8212;-(2)<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #ff0000;\">Maintenant, (1) &#8211; (2) \u00a0donne<\/span><\/p>\n<p style=\"text-align: justify;\">(T_{1}-T_{2})=(T_{2}+2\\Delta t-T_{1})V<\/p>\n<p style=\"text-align: justify;\">\n<span style=\"color: #ff0000;\">Faisant l&#8217;objet\u00a0V<\/span><\/p>\n<p>V=\\frac{((T_{1}-T_{2})v)}{T_{2}+2\\Delta t-T_{1}}<\/p>\n<p>V=\\frac{18-15}{(18+2.60-15).330} \\Rightarrow V=8,049m\/s<\/p>\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">V = 8,049 m\/s<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #ff0000;\">Maintenant, en rempla\u00e7ant V\u00a0dans l&#8217;\u00e9quation <span style=\"color: #3366ff;\">(1)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">\n<p>T_{1}v= 2D &#8211; T_{{1}}V &#8212;&#8212;(1)<\/p>\n<p>D=\\frac{T_{1}(V+v)}{2}<\/p>\n<p>D=\\frac{18(8,049+330)}{2} = 3046m<\/p>\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">D=3046m<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #ff0000;\"><span style=\"color: #3366ff;\"><strong>2)<\/strong><\/span> La distance minimale d au dessus de laquelle on n\u2019entend plus d\u2019\u00e9cho;<\/span><br \/>\n<span style=\"color: #000000;\">L\u2019\u00e9cho est entendu seulement si le temps mis par le son pour aller du paquebot \u00e0 la falaise et retourner est\u00a0\\geq 0,1s=dT<\/span>\u00a0, \u00a0<span style=\"color: #ff0000;\">d<\/span> <span style=\"color: #000000;\">correspondant \u00e0 la position A o\u00f9 le son mettra exactement 0,1s pour l\u2019aller et le retour.\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #ff0000;\">Donc:\u00a0VdT=2d-VdT <span style=\"color: #000000;\">\u00a0 \u00a0<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #ff0000;\"><span style=\"color: #000000;\">\u00a0\u00a0 \u00a0<\/span><\/span><\/p>\n<p>=&gt;\u00a0d=\\frac{(V+v)dT}{2}<\/p>\n<p>d=\\frac{(330+8,049).0,1}{2} = 17m<\/p>\n<p style=\"text-align: justify;\">\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">d=17m<\/span><\/span><\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"kamerpower-probleme-2-%e2%80%93-corrige-concours-ensp-yaounde-2015-chimie-et-physique\"><\/span><strong><span style=\"color: #3366ff;\">Kamerpower Probl\u00e8me\u00a02 &#8211;\u00a0<span style=\"color: #000000;\">Corrige Concours ENSP Yaounde 2015 Chimie et Physique\u00a0<\/span><\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Le <span style=\"color: #ff0000;\">pH<\/span> d\u2019une solution aqueuse d\u2019acide formique <span style=\"color: #ff0000;\">HCOOH<\/span> de concentration <span style=\"color: #ff0000;\">Co=2.10<sup>-2<\/sup> mol l<\/span><sup><span style=\"color: #ff0000;\">-1<\/span>\u00a0<\/sup>est \u00e9gal \u00e0 <span style=\"color: #ff0000;\">2,7<\/span>.<\/span><\/p>\n<ol style=\"text-align: justify;\">\n<li><span style=\"color: #000000;\">\u00a0Calculez la valeur du <span style=\"color: #ff0000;\">pKa<\/span> de cet acide.<\/span><\/li>\n<li><span style=\"color: #000000;\">\u00a0Combien de moles de formiate de sodium <span style=\"color: #ff0000;\">HCOONa<\/span> doit-on ajouter \u00e0 <span style=\"color: #ff0000;\">500 ml<\/span> de la solution initiale pour obtenir une solution de tampon de <span style=\"color: #ff0000;\">pH=3,7<\/span>\u00a0?<\/span><\/li>\n<li><span style=\"color: #000000;\">Combien de moles de soude doit-on ajouter \u00e0 la solution tampon de <span style=\"color: #ff0000;\">pH=3,7<\/span> pour obtenir une solution de formiate de sodium\u00a0? Quel est le <span style=\"color: #ff0000;\">pH<\/span> de cette solution\u00a0?<\/span><\/li>\n<li><span style=\"color: #000000;\">Tracez la courbe de neutralisation de l\u2019acide formique par la soude.<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"text-decoration: underline;\">N.B<\/span>\u00a0: On admettra que l\u2019addition des r\u00e9actifs n\u2019entra\u00eene pas de variation de volume.\u00a0On prendra <span style=\"color: #ff0000;\">log 2 = 0,3<\/span>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #ff0000;\">On utilisera les relations approch\u00e9es\u00a0:<\/span><br \/>\n<span style=\"color: #000000;\">Acide faible en solution (concentration molaire C)\u00a0:<span style=\"color: #ff0000;\"> pH=0,5. (pKa \u2013 log C)\u00a0;<\/span><\/span><br \/>\n<span style=\"color: #000000;\">\u00a0Sel d\u2019acide faible et de base forte en solution (concentration molaire C <strong>&#8216;<\/strong> )\u00a0:<\/span><br \/>\n<span style=\"color: #ff0000;\">pH = 7 + 0,5. (pKa + log C <strong>&#8216;<\/strong> ).<\/span><\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"kamerpower-solution-au-probleme-2-%e2%80%93-corrige-concours-ensp-yaounde-2015-chimie-et-physique\"><\/span><span style=\"color: #ff0000;\">Kamerpower Solution au probl\u00e8me 2 \u00a0&#8211;\u00a0<span style=\"color: #000000;\">Corrige Concours ENSP Yaounde 2015 Chimie et Physique\u00a0<\/span><\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"color: #3366ff;\"><strong>1)<\/strong> <\/span><span style=\"color: #000000;\">L&#8217;\u00e9quation de dissociation de l\u2019acide formique\u00a0\u00e0 l&#8217;\u00e9quilibre est:<\/span><br \/>\n<span style=\"color: #ff0000;\">HCOOH + H<sub>2<\/sub>O \u2192 HCOO<sup><span style=\"color: #000000;\">&#8211;<\/span> <\/sup>\u00a0+ H<sub>3<\/sub>O<span style=\"color: #000000;\"><sup>+<\/sup><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">Le pH d\u2019une solution de concentration <span style=\"color: #ff0000;\">C<sub>o <\/sub><\/span>de cet acide faible est donn\u00e9 par la formule\u00a0:<\/span><br \/>\n<span style=\"color: #ff0000;\">pH = 0,5 (pKa &#8211; logC<sub>o<\/sub>)<\/span><\/p>\n<p><span style=\"color: #000000;\">pKa = 2.pH + logC<sub>o<\/sub><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"color: #ff0000;\">Donc<\/span>: \u00a0pKa =\u00a02\u00a0.\u00a02, 27 + log (2 .10<sup>-2<\/sup>)\u00a0<\/span><\/p>\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">PKa = 3, 7<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"color: #3366ff;\"><strong>2)<\/strong><\/span> Le pH d\u2019une solution tampon est ;<\/span><\/p>\n<p>pH = pKa + log (baseconjugee\/acide)<\/p>\n<p style=\"text-align: justify;\">\n<p>pH = 3,7 + log (HCOO<span style=\"color: #000000;\"><sup>&#8211;<\/sup><\/span>\/HCOOH)<\/p>\n<p><span style=\"color: #000000;\">Le pH de la solution tampon <span style=\"color: #ff0000;\">[<\/span><span style=\"color: #ff0000;\">HCOO\u00a0<\/span><strong><span style=\"color: #ff0000;\"><sup>&#8211;<\/sup><\/span><\/strong><span style=\"color: #ff0000;\">]<\/span><strong><span style=\"color: #ff0000;\"> = <\/span><\/strong><span style=\"color: #ff0000;\">[<\/span><span style=\"color: #ff0000;\">HCOOH<\/span><span style=\"color: #ff0000;\">]<\/span><strong>\u00a0<\/strong>est <span style=\"color: #ff0000;\">3,7<\/span> ;\u00a0Il faut donc ajouter un nombre de moles de formiate de sodium \u00e9gal \u00e0 celui de l\u2019acide formique contenu dans les <span style=\"color: #ff0000;\">500 ml<\/span> de solution.<\/span><\/p>\n<p>Donc: 2 x 10<span style=\"color: #000000;\"><sup>-2<\/sup><\/span> x 0,5 = 10<span style=\"color: #000000;\"><sup>-2<\/sup><\/span> mole<\/p>\n<p style=\"text-align: right;\"><span style=\"color: #3366ff;\"><span style=\"text-decoration: underline;\">Nombre de moles =<\/span> 10<sup>-2<\/sup> <span style=\"text-decoration: underline;\">mole<\/span><\/span><\/p>\n<p><strong><span style=\"color: #3366ff;\">3)<\/span> <\/strong><span style=\"color: #000000;\">La totalit\u00e9 de la soude ajout\u00e9e r\u00e9agira avec l\u2019acide formique pour former \u00a0HCOONa selon la r\u00e9action;\u00a0<span style=\"color: #ff0000;\">HCOOH + NaOH\u00a0 \u2192 HCOONa + H<sub>2<\/sub>O<\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">Les <span style=\"color: #ff0000;\">500 ml<\/span> de solution tampon de <span style=\"color: #ff0000;\">pH = 3,7<\/span> contiennent <span style=\"color: #ff0000;\">10<sup>-2<\/sup> mole<\/span> d\u2019acide formique et <span style=\"color: #ff0000;\">10<sup>-2<\/sup> mole<\/span> de formiate de sodium. Donc,\u00a0pour neutraliser tout l\u2019acide formique, il faut ajouter\u00a0<span style=\"color: #ff0000;\">10<sup>-2<\/sup> moles<\/span> de soude,\u00a0donc,\u00a0<span style=\"color: #ff0000;\">2.10<sup>-2<\/sup> mole<\/span> de formiate dans <span style=\"color: #ff0000;\">500 ml<\/span> de solution, soit une concentration \u00a0<span style=\"color: #ff0000;\">C = 4.10<sup>-2<\/sup> mol l<sup>-1<\/sup><\/span>. \u00a0C\u2019est un sel d\u2019acide faible et de base forte<\/span><\/p>\n<p><span style=\"color: #000000;\">pH = 7 + 0,5.(pKa + logC) ,\u00a0dont le pH est celui d\u2019une base faible.<\/span><\/p>\n<p>pH = 7 + 0,5 x 3,7 + 0,5 x log(4&#215;10<span style=\"color: #000000;\"><sup>-2<\/sup><\/span>) = 8,15<\/p>\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">pH = 8,15<\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"color: #3366ff;\"><strong>4)<\/strong><\/span> Pour compl\u00e9ter la courbe de neutralisation, calculons le <span style=\"color: #ff0000;\">pH<\/span> apr\u00e8s addition d\u2019un nombre de moles de soude double de celui de l\u2019acide formique, soit <span style=\"color: #ff0000;\">2.10<sup>-2<\/sup> mole<\/span>. On a en solution <span style=\"color: #ff0000;\">10<sup>-2<\/sup> mole<\/span> de HCOONa<span style=\"color: #ff0000;\"> [de la neutralisation de 10<sup>-2<\/sup> de HCOOH]<\/span> et <span style=\"color: #ff0000;\">10<sup>-2<\/sup> mole<\/span> de NaOH en exc\u00e8s. Dans ce m\u00e9lange base faible\u00a0et\u00a0base forte, le pH est impos\u00e9 par la base forte. Donc la concentration de soude est\u00a0:<\/span><\/p>\n<p>C<span style=\"color: #000000;\"><sub>NaOH<\/sub><\/span> \u00a0= \u00a0(10<span style=\"color: #000000;\"><sup>-2<\/sup><\/span>x1000)\/500 = 2&#215;10<span style=\"color: #000000;\"><sup>-2<\/sup><\/span>\u00a0moll<span style=\"color: #000000;\"><sup>-1<\/sup><\/span><\/p>\n<p>pH = 14 + log(2&#215;10<span style=\"color: #000000;\"><sup>-2<\/sup><\/span>) = 12,3<\/p>\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">pH = 12,3<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u00a0Le diagramme se trouve ci-dessous;<\/span><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-4363\" src=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/01\/Kamerpower-1-chimie.png\" alt=\"R\u00e9vision gratuits pour le concours d\u2019entr\u00e9e \u00e0 l\u2019Ecole Nationale Sup\u00e9rieure Polytechnique (ENSP)\" width=\"522\" height=\"337\" srcset=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/01\/Kamerpower-1-chimie.png 522w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/01\/Kamerpower-1-chimie-300x194.png 300w\" sizes=\"auto, (max-width: 522px) 100vw, 522px\" \/><\/p>\n<p><strong>Recommandation<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/kamerpower.com\/fr\/concours-ensp-bamenda-3eme-annee-2019-2020-polytechnique-universite-de-bamenda\/\"><strong>Concours ENSP Maroua 3\u00e8me ann\u00e9e Ecole Nationale Sup\u00e9rieure Polytechnique.<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/kamerpower.com\/fr\/concours-professionnel-information-2015-2016-minfopra-recrutement-de-personnels-dans-le-corps-des-fonctionnaires-de-linformation-kamerpower-cameroun\/\"><strong>Concours Professionnel Information MINFOPRA recrutement de personnels Cameroun<\/strong><\/a><\/li>\n<\/ul>\n<p style=\"text-align: center;\"><span style=\"color: #3366ff;\"><strong>Corrige Concours ENSP Yaounde 2015 Chimie et Physique &#8211;\u00a0Ecole Nationale Polytechnique de Yaound\u00e9<\/strong><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5914 size-full\" src=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP.jpg\" alt=\"ENSP Concours R\u00e9vision Cameroun\" width=\"1280\" height=\"720\" srcset=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP.jpg 1280w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-600x338.jpg 600w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-300x169.jpg 300w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-768x432.jpg 768w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-720x405.jpg 720w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-520x293.jpg 520w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-320x180.jpg 320w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-150x84.jpg 150w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-1200x675.jpg 1200w\" sizes=\"auto, (max-width: 1280px) 100vw, 1280px\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":5914,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_themeisle_gutenberg_block_has_review":false,"footnotes":""},"categories":[1701],"tags":[1712,1727,2078],"class_list":["post-12238","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-epreuves","tag-epreuve-de-concours-dentree-aux-grandes-ecoles-cameroun","tag-epreuve-ensp-concours-cameroun","tag-sujets-corrigees-concours-ensp-yaounde-polytechnique"],"_links":{"self":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts\/12238","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/comments?post=12238"}],"version-history":[{"count":1,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts\/12238\/revisions"}],"predecessor-version":[{"id":67821,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts\/12238\/revisions\/67821"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/media\/5914"}],"wp:attachment":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/media?parent=12238"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/categories?post=12238"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/tags?post=12238"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}