ssNake's Chemistry [u2] Q's

[REBORN]

haha that's because it's [u3]

well I was wondering why the answers say step(2)is:

"mix the sample with water to dissolve Cl- ions" ---> y not Na+ ions? and what's the purpose?

[HenryP]

I assume during one of the steps it tells you to add Barium Sulfate to the solution? The reason we want to dissolve the Chloride ions is so that we can precipitate them as Barium Chloride which can then be weighed. The Sodium ions don't really serve a purpose in the gravimetric analysis.

[REBORN]

Ah I think I get it.

--> weigh the sample
--> mix the sample with water to dissolve Cl- ions
--> filter the mixture to get rid of the water and you're left with the savoury spread with dissolved Cl- ions? (is that right?)
--> form the precipitate
--> filter the precipitate and wash with water -- why?
--> weigh precip.

[Lasercookie]

When we prepared a standard solution of Sodium Carbonate in class (volumetric analysis), we washed the filtered stuff with ethanol, to ensure that there were no impurities in the substance. 

I suspect this would have a similar purpose.

[REBORN]

Okay.

Q: A 0.693 g sample of a silver alloy used to make cutlery is dissolved completely in nitric acid. Excess sodium chloride solution is added to produce a precipitate of silver chloride. The precipitate is filtered, dried and found to weigh 0.169 g.

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I'm having trouble realising/forming the full equation for what's going on here....

[stonecold]

The Nitric acid is just used to dissolve the silver i.e. Ag (s) --> Ag+ (aq)

Then, you just use the equation Ag+ (aq) + Cl- (aq) --> AgCl (s)  to solve. 

[REBORN]

I know that's the ionic but I like writing the full so I can picture what's going on...

What is the full one?

ty.

[stonecold]

That is all that is going on.  Instead of using water as the solvent, you are using acid to dissolve the silver, that's all. 

Then the full ionic equation is:

Na+ (aq) + Cl-(aq) + Ag+(aq) --> Na+(aq) + AgCl(s)

The sodium ions are spectator ions.  They don't undergo any reaction.

[REBORN]

Okay.

Q: A 0.693 g sample of a silver alloy used to make cutlery is dissolved completely in nitric acid. Excess sodium chloride solution is added to produce a precipitate of silver chloride. The precipitate is filtered, dried and found to weigh 0.169 g. Find the % of silver in the alloy

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I'm having trouble realising/forming the full equation for what's going on here....

So I got the answer to be 18.4%.

Part b asks "If the precipitate was not completely dry when weighed, what effect would this have on the answer for part a?"

[vea]

If the precipitate was not completely dry when weighed, it is heavier than it would be if it was completely dried. This means that you wold get a higher mass of precipitate and a higher % of silver in the alloy.

[REBORN]

Say you have 1g of precip with 0.5g Fe and 0.5g Z in it (hypothetically)

If you had the mixture completely dry you'd get 50% Z
If you had the mixture at 1.2g with say water adding the 0.2g - you'd have 0.5/1.2 = 41.6% Z

Am I missing something here? It seems the higher mass of precip decreases the % of the silver?? :S

[vea]

The precipitate isn't the alloy, it's just silver chloride. This means that the mass of the precipitate is directly proportional to the % of silver in the alloy.

[REBORN]

During the preparation of the standard solution why is water added to the level of the calibration mark on the flask after the solid has dissolved, rather than before?

[Lasercookie]

During the preparation of the standard solution why is water added to the level of the calibration mark on the flask after the solid has dissolved, rather than before?

Remember that molarity is the amount of solute dissolved into a volume.

You may have also noticed that when you dissolve stuff, the water level goes down a bit. (I can't remember the exact reason for this).

If we added the water in before, the water level would go down and we wouldn't know much about the concentration of the standard solution (which is vital for the later calculations.).

[REBORN]

A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus:

a the pipette used to deliver the aliquot of sodium carbonate solution
b the flask containing the aliquot
c the burette.

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I'm interested in why for each ans.

ty!