Hey guys, can someone let me know if I'm doing this right?
2MnO4 + 16H+ +5(C2O4)2 --> 2Mn(2+) +8H2O +10CO2
use this equation to calculate the volume of a manganate(VII) solution of concentration 0.0200 M, required for complete reaction with 0.2000g of iron (II) oxalate.
This is what I did:
n(FeC2O4)= 0.2000/143.8 = 0.00139 mol
n(MnO4) = 5/2 x 0.00139 = 0.00348 mol
therefore, v(MnO4)= 0.00348/0.0200 = 0.174 L
looks good, except watch the sig figs. the answer to the first equation should be to 4 sig figs. i'd also add more working. for example after i calculate n(FeC2O4) i'd state explicitly that the answer to that equals to n(C2O4). notice that iron(II) oxalate isn't actually present in the equation provided. i would also watch the use of superscripts. permanganate ion has a charge of 1-, so you should include that in your working.
EDIT: and yes, the 2/5 thing as pointed out by rod.
When the indicator permanently changes colour, doesn't that mean the acid and base that react are stoichiometrically equivalent?
I always thought the end point was the point the indicator first changes colour? (not permanent)
And the equivalence point where the indicator changes colour permanently. :/
Someone help distinguish the two please 
you'll find that sometimes the solution in the conical flask into which you are dispensing the solution in the burette will change colour at the first drop. the colour will of course quickly vanish, but a colour changes occurs nevertheless. it is a permanent colour change that we are looking for. a pH indicator is a weak acid/base whose conjugate base/acid is of a different colour. we exploit this property during titration to determine roughly where the equivalence point lies. envisage this. say you have a solution of HCl in the conical flask and a solution of NaOH in the burette, and you add a few drops of phenolphthalein into the HCl. phenolphthalein is itself a weak acid, which has the formula C20H14O4, but for the purposes of this explanation, we shall suppose that the formula is simply HIn instead. now the moment you add the phenolphthalein into the HCl solution (which, remember, is basically just a bunch of H+ and Cl- ions floating around amidst a sea of H2O molecules), the phenolphthalein molecules will react with the water according to the equation:
HIn(aq) + H2O(l) <--> In-(aq) + H3O+(aq)
no colour pink
the concentration fraction for this reaction is therefore:
CF = [In-][H3O+]/[HIn]
when the reaction reaches equilibrium, the concentration fraction becomes the equilibrium constant, which for phenolphthalein is 7.9*10^(-12) according to google.
Ka = [In-][H3O+]/[HIn] = 7.9*10^(-12)
now you can see that once the reaction reaches equilibrium, there is A LOT more HIn floating around than there are In-. this is why HIn is the dominant species, and the overall solution appears colourless initially. but as we add more and more NaOH into the solution, there will be less and less H3O+. so to regain equilibrium, the system will favour the forward reaction so as to replenish the lost H3O+ ions. this means less HIn and more In-. eventually, we'll get to a stage where In- becomes the dominant species, at which point the solution will appear pink.
but when exactly does the transition from colourless to pink occur, you may ask? i said earlier that the first colour changes is often experienced at the first drop. this is mainly because when the first drop hits the solution, in the region where the drop hits, there will be a sudden deficiency of H3O+ ions, and the system will immediately respond by favouring the forward reaction, producing more In- at the expense of HIn. but all of this happens locally and almost instantaneously. when the OH- disperse, the deficiency becomes less pronounced, and so the solution quickly returns to colourless. but there is a point where the overall deficiency of H3O+ is so high that the colour change is more or less permanent (permanent is a dodgy word to use here because you'll find that sometimes if you leave the solution out for a few hours it will go back to being colourless). when does exactly does this occur? to appreciate this, we must first do a little maths.
let us return to the equilibrium law expression and do a little bit of rearranging:
Ka = [In-][H3O+]/[HIn]
pKa = -log_10([In-][H3O+]/[HIn])
pKa = -log_10[H3O+] - log_10([In-]/[HIn]) (yay log laws from methods)
pKa = pH - log_10([In-]/[HIn])
pH = pKa + log_10([In-]/[HIn])
this last equation is what we call the henderson-hasselbalch equation. fancy name for a not-so-fancy equation. it so happens that the human eye (which btw is a very bad detector) can detect a colour change only when the ratio [In-]/[HIn] is between 0.1 and 100. this means that the human eye can only detect a colour change when the pH of the solution is:
pH = pKa +- 1 (where +- is the plus/minus symol)
now the Ka of an indicator varies according to how weak an acid/base it is, but we know that for phenolphthalein, Ka = 7.9*10^(-12) as stated above. this means that pKa = 9.7. so this means that the human eye can detect a colour change only when the pH of the solution is between 8.7 and 10.7. this is basically how they determine the pH range of a particular indicator. everything ties back to Ka values.
how do we choose which indicator to use? well, we know for instance that if we react a strong base with a strong acid, the reaction will reach completion (not equilibrium because the species are strong) roughly when pH = 7 (and you can prove why this is the case). so we want an indicator that will tell us when the pH of the solution is roughly 7; the pH curve is very steep around the pH = 7 region, so if we are off by a bit it doesn't really matter, and it so happens that phenolphthalein can do this because the human eye detects a colour change in phenolphthalein when pH is between 8.7 - 10.7, as per all the calculations above (you might think that this is way off pH = 7, but if you look at the standard pH curve for strong acid/strong base reaction, you'll find that it is not really).
man didn't envisage this post to be so long but hopefully i've shed some light on how indicators work...
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