http://weewee.sgforums.com/forums/2297/topics/323144 en-US 60 <p>hi thanks all for help.. hi jiaxing2.. an aqueous solution do contain H+ and OH- ions.. dissociated from water molecules.</p> <p>thanks <img src="/images/emoticons/kde-3.5.8/set1/wink.png" alt= "wink.png" /></p> Sun, 06 Jul 2008 01:26:49 +0800 weewee.sgforums.com:2297:323144:8221139 arigatoast http://weewee.sgforums.com/forums/2297/topics/323144 <p>The ions present are H+, OH-, Na+, Cl-. You have to consider the standard electrode/reduction/oxidation potentials for the reduction and oxidation half equations for the 4 ions.</p> <p>&nbsp;</p> <p>Between the cations, there is no contest. Na+ is too stable (since Na is too reactive) to be reduced, so 2H+ will be reduced to H2 gas.</p> <p>&nbsp;</p> <p>Between the anions, you will have to factor in the molarities. Here, the molarity of Cl- is high enough (notice that standard reduction potentials given in the data booklet&nbsp;are for standard conditions, ie. 1 mol/dm3; here we have 5 mol/dm3)&nbsp;such that Cl- will be certainly be preferentially oxidized instead of OH-.</p> <p>&nbsp;</p> <p>As H+ is reduced and Cl- is oxidized, what is left over will be NaOH sodium hydroxide solution, as <a href= "http://www.sgforums.com/users/241361" rel= "nofollow">arigatoast</a>&nbsp;predicted.</p> <p>&nbsp;</p> <p>Since NaCl is a neutral salt (product of strong acid and strong alkali), we know the molarity of [H+] to begin with (Note that pH = 7 is neutral only under 25 deg C; the question doesn't state the temperature and pressure, so we assume it's 25 deg C and 1 atm).</p> <p>&nbsp;</p> <p>Based on the electrons transferred (see <a href= "http://www.sgforums.com/users/202051" rel= "nofollow">secretliker</a>'s post above), we can work out exactly how much H+ is removed from solution. Hence we can work out the new no. of mol of&nbsp;H+ and hence the&nbsp;new&nbsp;[H+]&nbsp;of solution after the current is passed, and therefore pH.</p> <p>&nbsp;</p> <p>Note : Usually for alkaline solutions,&nbsp;we first&nbsp;calculate pOH, and hence obtain pH. In this case, however, we're dealing directly with the reduction half equation that removes protons H+, so no problem here.</p> <p>&nbsp;</p> <p>&nbsp;</p> Sat, 05 Jul 2008 10:46:55 +0800 weewee.sgforums.com:2297:323144:8219323 UltimaOnline http://weewee.sgforums.com/forums/2297/topics/323144 <p>a solution of Nacl does nt cotain H and OH ions....</p> <p>i think what happens at the cathode is 2 H2O + 2e ---&gt; 2OH- + H2</p> <p>pOH=2.26 then pH shld be 11.74 ba...</p> <p>correct me if im wrong...i 2 yrs nv touch chem at all...</p> <p>&nbsp;</p> <p>&nbsp;</p> Sat, 05 Jul 2008 08:43:32 +0800 weewee.sgforums.com:2297:323144:8219141 jiaxing2 http://weewee.sgforums.com/forums/2297/topics/323144 <p>Q = It</p> <p>Q = 0.25 (35*60) = 525 Coulombs</p> <p>Number of moles of electrons passed</p> <p>= 525 / Faraday Constant</p> <p>= 525 / 96500 = 5.4404 * 10^-3 mol</p> <p>pH = -lg(5.4404*10^-3) = 2.26? (I think I'm wrong.)</p> Sat, 05 Jul 2008 02:53:20 +0800 weewee.sgforums.com:2297:323144:8219017 secretliker http://weewee.sgforums.com/forums/2297/topics/323144 <p>a current 0.25A&nbsp; is passed thru&nbsp; 400 cm3 of a 5 M&nbsp; solution of NaCl&nbsp; for 35 mins ,using C electrodes.&nbsp; What will be the pH&nbsp; of the solution after&nbsp; the current is turned off?</p> <p>&nbsp;</p> <p>what is tink is H and Cl ions will be discharged so what is left will be an alkaline solution, but i'm not sure how to go abt calculating the pH... i do not have the answer to this qns.</p> <p>&nbsp;</p> <p>any help is much appreciated. thanks.</p> Sat, 05 Jul 2008 02:32:15 +0800 weewee.sgforums.com:2297:323144:8218994 arigatoast http://weewee.sgforums.com/forums/2297/topics/323144