Below is a list of substances
NaOH
AgCl
PbO
Mg(OH)2
CaCO3
(NH4)2SO4
MgCl2
Dilute HNO3
CaO
Pb(NO3)2
2 substances when reacted in the aqueous state forms a white precipitate that dissolves when one of the reactants is added in excess.
My answer is NaOH and (NH4)2SO4, these are the reactants.
Is it correct?
Originally posted by anpanman:Below is a list of substances
NaOH
AgCl
PbO
Mg(OH)2
CaCO3
(NH4)2SO4
MgCl2
Dilute HNO3
CaO
Pb(NO3)2
2 substances when reacted in the aqueous state forms a white precipitate that dissolves when one of the reactants is added in excess.
My answer is NaOH and (NH4)2SO4, these are the reactants.
Is it correct?
The white ppt is lead(II)hydroxide, which reacts with excess hydroxide ions to form soluble tetrahydroxoplumbate(II) ions.
Originally posted by UltimaOnline:
The white ppt is lead(II)hydroxide, which reacts with excess hydroxide ions to form soluble tetrahydroxoplumbate(II) ions.
Hi,
May I know if the reaction between Na3N and HCl produces NH3 and NaCl? It looks like it but in my question, they said that the reaction produces solution S. And S, when boiled with NaOH, gives a gas G which turns damp litmus from red to blue.
Name the salt present in S that reacts with aq NaOH to form gas G. I think it is the NO3- ions since this test is for presence of NO3- ions. But the products don't seem to have any compound that has the ion. Thanks.
Originally posted by anpanman:
Hi,May I know if the reaction between Na3N and HCl produces NH3 and NaCl? It looks like it but in my question, they said that the reaction produces solution S. And S, when boiled with NaOH, gives a gas G which turns damp litmus from red to blue.
Name the salt present in S that reacts with aq NaOH to form gas G. I think it is the NO3- ions since this test is for presence of NO3- ions. But the products don't seem to have any compound that has the ion. Thanks.
1st step is (tetra)protonation of the trinegative nitride anion N3- (4 lone pairs, no bond pairs), easily achieved since all sodium salts are completely soluble / strong electrolytes : :
N3- + 4H+ --> NH4+
The +ve charges from 3 Na+ cations and 1 NH4+ cation, are counterbalanced by the 4 Cl- ions.
Na3N + 4HCl --> NH4Cl + 3NaCl
A mixture of NaCl(aq) and NH4Cl(aq) is obtained (ie. Na+, NH4+ and Cl- ions present).
Proton transfer (ie. Bronsted-Lowry acid-base reaction) occurs with aqueous hydroxide ions.
Ionic :
NH4+ + OH- --> NH3 + H2O
Chemical :
NH4Cl + NaOH --> NH3 + H2O + NaCl
Gaseous ammonia undergoes hydrolysis on the moist red litmus paper to produce OH- ions which turns red litmus blue :
NH3 + H2O --> NH4+ + OH-
Notice that this is exactly the reverse of the earlier proton transfer reaction (see above).
Originally posted by UltimaOnline:
1st step is (tetra)protonation of the trinegative nitride anion N3- (4 lone pairs, no bond pairs), easily achieved since all sodium salts are completely soluble / strong electrolytes : :
N3- + 4H+ --> NH4+
The +ve charges from 3 Na+ cations and 1 NH4+ cation, are counterbalanced by the 4 Cl- ions.
Na3N + 4HCl --> NH4Cl + 3NaCl
A mixture of NaCl(aq) and NH4Cl(aq) is obtained (ie. Na+, NH4+ and Cl- ions present).
Proton transfer (ie. Bronsted-Lowry acid-base reaction) occurs with aqueous hydroxide ions.
Ionic :
NH4+ + OH- --> NH3 + H2O
Chemical :
NH4Cl + NaOH --> NH3 + H2O + NaCl
Gaseous ammonia undergoes hydrolysis on the moist red litmus paper to produce OH- ions which turns red litmus blue :
NH3 + H2O --> NH4+ + OH-
Notice that this is exactly the reverse of the earlier proton transfer reaction (see above).
Hmm, quite clear explanations. Thank you. I have 2 questions. First, if we have 2 reactants and we are supposed to write down the reaction (including its product), how do we know what the products are if it is some kind of foreign reactions which we have never seen before?
2. If we are doing electrolysis, with the electrolyte being aq. H2SO4 and the electrodes being aluminium, you will see that there is an initial increase in mass at the Al anode but the electrolysis stops shortly after.
As to why there is an increase, could it be due to the formation of Al(OH)3? Or is it Al2(SO4)3? And the reaction stops because of the presence of Al(OH)3 or Al2(SO4)3 surrounding the anode thus causing no more anions to be discharged at anode? Sorry, but I am not sure of whether the above 2 aluminium salts are soluble... tried to check up but can't find much reliable sources.
And by the way, apologise for having so many questions to ask. ;p
1) Normally if a reaction does happen with unknown substances, just swop the ions around to get the new products or use similar compounds that you have studied to guess the product. I don't really think O's will really give you something you totally dunno for a question.
2) Not to sure about this but Al(OH)3 would react with the acid since its amphoteric, Al2(SO4)3 should not react since it cannot form any product with H2SO4
If a layer really does form, it should be Al2(SO4)3 and the reason the reaction stop is that the layer of salt acts as a insulator and prevents further exchange of e- between the metal the solution.
Both Al(OH)3 and Al2(SO4)3 are relatively insoluble in pure water, and may form precipitates.
http://en.wikipedia.org/wiki/Al(OH)3
http://en.wikipedia.org/wiki/Al2(SO4)3
However, as dkcx pointed out, since Al(OH)3 is amphoteric (since it has both covalent and ionic character, due to the high charge density of Al3+ ion), it is soluble in (ie. it can react to form soluble compounds with) both acids and alkalies.
Which begets the next question : would the electrolyte solution be acidic or alkali as the electrolysis carries on? There are a few factors/processess responsible for H+ (or H3O+) and OH- concentration in the electrolyte :
1) H+ (or H3O+) from H2SO4 (the original electrolyte).
2) OH- from H2O as H+ (or H2O) is reduced at the cathode.
3) H+ (or H3O+) from the hydrolysis of Al3+ cation (high charge density on Al3+ polarizes electron density of the O-H bond in water ligands, allowing for ready proton dissociation).
So the pH of the solution at any given time, depends on several factors. Without further data (including the molarity of the H2SO4 acid used as the initial electrolyte) and further calculations (beyond the scope of both 'O' and 'A' levels), it is impossible to say for sure which, or how much of each, insoluble precipitate, will be present.
This is not a fair questions (at either 'O' or 'A' levels), and so you shouldn't encounter such questions in your exam. However, because of its amphoteric nature (ie. it can react with both acids and alkalis to form soluble compounds). it is a safer bet to assume, as dkcx did, that the ppt is Al2(SO4)3 rather than Al(OH)3 (which could also be present in some quantity at some stage, before being reacted/dissolved away).
And both anpanman and dkcx have partially (but incompletely) identified the reason why the ppt (whichever it may be) stops the current from flowing - the buildup of positive charge at the anode (now coated with a ppt that acts as a physical barrier) prevents electrons from flowing from the anode to the electric cell to the cathode.
anpanman said : >>> ppt surrounding the anode thus causing no more anions to be discharged at anode <<<. Are you sure anions from solution are discharged? If so, what is the anion? Try focusing on the (reactive) electrode instead (the passivation layer of Al2O3 must first be removed for the Al metal pieces to function as electrodes). And remember - Red Cat riding An Ox (Reduction at Cathode, at Anode Oxidation occurs).
dkcx said : >>> the reason the reaction stop is that the layer of salt acts as a conductor and prevents further exchange of e- between the metal the solution. <<< "Acts as a conductor"? Solid precipitates cannot conduct electricity, as the ions are immobile. "Exchange of e- between metal and solution"? That takes place at the cathode, not anode. According to anpanman's question, the ppt coats the anode (which makes sense since that's where Al3+ ions are generated); and at the anode, e- is transferred from the metal to the electric cell to the cathode, not to (any ions in) the solution.
Opps, i meant insulator not conductor... Paisei
Hi again.
I want to clarify with UltimaOnline (although others can still help) regarding the electrolysis question you have explained. If you want to electrolyse H2SO4 and you have the reactive electrode (say, Al electrode) then will electrolysis go on? If yes, then the thing being electrolysed is the Al part right? From Al to Al3+ at the anode?(which means Al3+ ions are being discharged?)
Do reply asap. Appreciate all your help.
Originally posted by anpanman:Hi again.
I want to clarify with UltimaOnline (although others can still help) regarding the electrolysis question you have explained. If you want to electrolyse H2SO4 and you have the reactive electrode (say, Al electrode) then will electrolysis go on? If yes, then the thing being electrolysed is the Al part right? From Al to Al3+ at the anode?(which means Al3+ ions are being discharged?)
Do reply asap. Appreciate all your help.
"Discharge" means to remove the charge.
Hence, discharge of cations (eg. H+ ions) will be reduction (which takes place at the cathode), and discharge of anions (eg. Cl-) will be oxidation (which takes place at the anode).
Therefore, when Al is oxidized to Al3+ at the anode, you are creating a charge, not removing a charge (or "discharging").
Also remember that 'electrolysis' means you are providing a battery, and how you choose to orientate the battery will determine which electrode is the anode, and which electrode is the cathode.
So in your setup,
Al will be oxidized to Al3+ at the anode.
H+ will be reduced to H2 gas at the cathode.
Originally posted by UltimaOnline:
"Discharge" means to remove the charge.
Hence, discharge of cations (eg. H+ ions) will be reduction (which takes place at the cathode), and discharge of anions (eg. Cl-) will be oxidation (which takes place at the anode).
Therefore, when Al is oxidized to Al3+ at the anode, you are creating a charge, not removing a charge (or "discharging").
Also remember that 'electrolysis' means you are providing a battery, and how you choose to orientate the battery will determine which electrode is the anode, and which electrode is the cathode.
So in your setup,
Al will be oxidized to Al3+ at the anode.
H+ will be reduced to H2 gas at the cathode.
Thanks. By the way, if the quesiion asks me for "which salt can be prepared through titration?"
Can I write NaOH? I know (NH4)2SO4 can surely be obtained through titration. But is NaOH considered a salt since it is an alkali? thanks
We don't call NaOH a salt.
Any salt that is form in a reaction between an acid and a base can be obtained by tiration such as NaCl, Na2SO4 etc etc