25 ml of 10% solution is taken and added into 50 ml waterFurther 10ml Bacl2 (20%) is added in itIt will make milky solution,add some drops of phenophthaein indicator, heat it to boiling point so as to remove chlorine from it Now titrate it against 0.25N (NaoH)Measure the volume at which the milky solution turns into pinkish Multiply the volume with 0.1696 and this will give %age alumina in aluminium sulphate.PLEASE CHECK WHATS WRONG WITH THIS PROCEDUREOR IF HAVE SOME OTHER METHOD TO CHECK ALUMINA PLEASE HELP.
On a blanket or upholstery, off white to light graySome other materials might produce a darker smoke.
There are a few things that don't sound right in your explanation: First of all, refluxing the solution won't get rid of chloride ions and you certainly don't have a strong oxidant to be forming chlorine gasYou should scratch this explanation from your report because it makes no senseSecond of all, when BaSO4 precipitates to leave behind aluminum, the Lewis acidity of that ion is strong enough to undergo multiple deprotonationsIn fact, in neutral or slightly alkaline solutions you end up forming Al(OH)3: [Al(H2O)6]3+(aq) - 3 H3O+(aq) + Al(OH)3(s) The heating is only to ensure completion of this processThird of all, phenolphthalein changes color when the solution changes from acidic to basic (pH ~ 8.0)Therefore, NaOH neutralizes all three equivalents of acid produced by aluminum before the indicator turns pink: 3 H3O+(aq) + 3 NaOH(aq) - 3 Na+(aq) + 3 H2O(l) what this means is that to get the moles of aluminum present in solution, you have to divide the number of moles of NaOH by 3.
