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Topic: Simple Equation  (Read 18460 times)

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crcchelsea

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Simple Equation
« on: April 05, 2005, 09:20:56 PM »
I need an example equation result to make sure I'm doing it right...I'll give the reaction if u can give me the result.


200ml each of .5 M Na2S (aq) and .45 M FeCl3 (aq) ----->   ?
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crcchelsea

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Re:Simple Equation
« Reply #1 on: April 05, 2005, 09:33:16 PM »
OK, i think i got the result but i can't balence it!!!   its:


NaCL  +  FeS  right?
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crcchelsea

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Re:Simple Equation
« Reply #2 on: April 05, 2005, 09:48:34 PM »
I think I'm answering  my own questions here but can someone tell me if this is right?


3Na2S(aq)  +  2FeCl3(aq) -------> 6NaCl (s)  +  Fe2S3(aq)

and the net ionic equation is:

6Na+ (aq)  + 6Cl-(aq) -----> 6NaCl (s)

Is that all right???
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Offline Donaldson Tan

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Re:Simple Equation
« Reply #3 on: April 05, 2005, 11:01:46 PM »
on first thought, it might seem the reaction might be like that. However, sulphide ions hydrolyses readily to produce H2S, so the actual reaction mechanism is:

1. Na2S + 2H2O -> 2NaOH + H2S
2. 3NaOH + FeCl3 -> Fe(OH)3 + 3NaCl

so the overall reaction is:
3Na2S + 6H2O + 2FeCl3 -> 2Fe(OH)3 + 6NaCl + 3H2S

since water is always excess in an aqueous system, we need not consider it to be the limiting reagent. From the above equation, we can tell that 1 mole of FeCl3 reacts with 1.5moles of Na2S.

given we are considering 200ml 0.5M Na2S and 200ml 0.45M FeCl3 whereby there are 0.1mole of Na2S and 0.09mol of FeCl3, we can conclude FeCl3 is in excess and thus Na2S is the limiting reagent. Being the limiting reagent, all Na2S is consumed during the reaction. This should be the basis of your calculation for the number of moles of reactant and product present after the reaction has completed.
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Offline Borek

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Re:Simple Equation
« Reply #4 on: April 06, 2005, 06:39:11 PM »
Quote from: geodome on April 05, 2005, 11:01:46 PM
so the overall reaction is:
3Na2S + 6H2O + 2FeCl3 -> 2Fe(OH)3 + 6NaCl + 3H2S

You are probably right, but it can be interesting to check another possibility:

2Fe3+ + S2- -> S + 2Fe2+

Hydrolysis of S2- and Fe(OH)3 precipitation probably shifts the potentials so that this reaction is not possible, but at first sight Fe3+ is strong oxidizing agent and S2- is easily oxidized to S...
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Offline AWK

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Re:Simple Equation
« Reply #5 on: April 07, 2005, 04:04:10 AM »
Na2S and FeCl3 produce Fe2S3 (black) . This compound hydrolyses in water slowly (much faster in hot water) to form Fe(OH)3.
In common qualitative analysis procedure, where H2S is used in acidic medium before precipitation of Fe with (NH4)2S, Fe(3+) is reduced by H2S to Fe(2+).
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Garneck

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Re:Simple Equation
« Reply #6 on: April 07, 2005, 02:05:05 PM »
Quote from: geodome on April 05, 2005, 11:01:46 PM
1. Na2S + 2H2O -> 2NaOH + H2S

No way that would happen. You've got yourself a base and an acid in your products. That reaction would go the opposite way..
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Offline Borek

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Re:Simple Equation
« Reply #7 on: April 07, 2005, 06:42:20 PM »
Quote from: Garneck on April 07, 2005, 02:05:05 PM
No way that would happen. You've got yourself a base and an acid in your products. That reaction would go the opposite way..

Not exactly - there will be very strong hydrolysis. pH of 0.01M Na2S solution is 11.77. So instead of dissolving 0.01M Na2S you can dissolve 0.006M NaOH to obtain the same pH.
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Offline Donaldson Tan

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Re:Simple Equation
« Reply #8 on: April 07, 2005, 07:21:48 PM »
H2S is a very weak acid.
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Offline Donaldson Tan

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Re:Simple Equation
« Reply #9 on: April 07, 2005, 07:23:28 PM »
Quote from: AWK on April 07, 2005, 04:04:10 AM
Na2S and FeCl3 produce Fe2S3 (black) . This compound hydrolyses in water slowly (much faster in hot water) to form Fe(OH)3.
In common qualitative analysis procedure, where H2S is used in acidic medium before precipitation of Fe with (NH4)2S, Fe(3+) is reduced by H2S to Fe(2+).

i think AWK is most probably right

 :)
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Garneck

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Re:Simple Equation
« Reply #10 on: April 08, 2005, 02:33:31 AM »
Quote from: Borek on April 07, 2005, 06:42:20 PM
Not exactly - there will be very strong hydrolysis. pH of 0.01M Na2S solution is 11.77. So instead of dissolving 0.01M Na2S you can dissolve 0.006M NaOH to obtain the same pH.

I'm not going to argue with numbers.. but common sense tells me that NaOH will react with H2S. So the equlibrium of that reaction would be moved to the left - forming of Na2S.
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Offline Borek

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Re:Simple Equation
« Reply #11 on: April 08, 2005, 05:10:39 AM »
Quote from: Garneck on April 08, 2005, 02:33:31 AM
I'm not going to argue with numbers.. but common sense tells me that NaOH will react with H2S. So the equlibrium of that reaction would be moved to the left - forming of Na2S.

Don't believe in common sense ;) Believe in numbers.

Only in case of very simple systems containing strong bases and strong acids reactions are simple. If you have a weak acid and/or weak base in the solution, you have always a mixture of all possible forms. Usually some of the forms are dominating. In case of solutions of sulfides S2- is never a dominant form, unless you add large excess of strong hydroxide to solution.
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Garneck

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Re:Simple Equation
« Reply #12 on: April 08, 2005, 07:07:43 AM »
Quote from: Borek on April 08, 2005, 05:10:39 AM
Don't believe in common sense ;) Believe in numbers.

Only in case of very simple systems containing strong bases and strong acids reactions are simple. If you have a weak acid and/or weak base in the solution, you have always a mixture of all possible forms. Usually some of the forms are dominating. In case of solutions of sulfides S2- is never a dominant form, unless you add large excess of strong hydroxide to solution.

Well in this case, we'll have a strong base cation and a weak acid cation. Since the compund is ionic, it will remain in dissociated form and the high pH will be caused by the presence of Na+ (which causes hydrolisis of water and so on and so on). Ok? ;)
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Offline Borek

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Re:Simple Equation
« Reply #13 on: April 08, 2005, 08:42:25 AM »
Quote from: Garneck on April 08, 2005, 07:07:43 AM
Well in this case, we'll have a strong base cation and a weak acid cation. Since the compund is ionic, it will remain in dissociated form and the high pH will be caused by the presence of Na+ (which causes hydrolisis of water and so on and so on). Ok? ;)

No.

What is a 'weak acid cation'?

Presence of Na+ can only lower pH - and in fact it does, but the effect is very low. pKb for NaOH is 0.2. Calculated pH of 0.1M NaCl solution is 6.97 (or 6.98 if you take activities into account) due to the Na+ hydrolysis.
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Garneck

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Re:Simple Equation
« Reply #14 on: April 08, 2005, 08:50:47 AM »
Quote from: Borek on April 08, 2005, 08:42:25 AM
No.

What is a 'weak acid cation'?

Presence of Na+ can only lower pH - and in fact it does, but the effect is very low. pKb for NaOH is 0.2. Calculated pH of 0.1M NaCl solution is 6.97 (or 6.98 if you take activities into account) due to the Na+ hydrolysis.

Sorry, I ment to say "weak acid anion" - meaning S2- (anion from a weak acid).

You've just chosen a bad example for calculations. NaCl is a salt of a strong acid and a strong base. That means the salt solution will have a neutral pH (meaning pH = 7). You wrote that the calculated pH=6,98 so we can safely say it's neutral.

Try calculating pH for.. for example Na2SO3. You'll get a high pH, because it's a salt of a weak acid and a strong base.
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