Calculate the amount concentration of silver nitrate lab?

250mL of silver nitrate is poured into a beaker, 20cm coil wire is placed in solution,(left for a couple days)
Once the coil wire is removed the silver crystals remain in the beaker.
The mass of a filtered piece of paper is .94g
Then the beaker is filtered to seperate the solid silver
The mass of the dry silver is 2.10g

How do I find the amount concentration? and next question im stuck on is, Describe a test that could be performed to ensure that all silver ions from the solution react to form a precipitate. Indicate what would be observed in a positive result for the presense of silver ions and what would be observed in a negative result.

thanks

It’s been a while since I had to do these, but here are my thoughts. Assume mass of silver is without filter paper (check if this is the case).

Your eqn is going to be:

Cu(0) + 2AgNO3 -> Cu(NO3)2 + 2Ag(0)

Concentration problems consist of a few steps.

1. Find the amount of elemental silver product in mols?

n = m/M = 2.10g / 107.8682 g/mol
n = 1.947×10^-2 mol

2. What is the vol of solution?

V = 0.250 L (given)

3. Molarity

[AgNO3] = n/V = 1.947×10^-2 mol / 0.250 L
[AgNO3] = 7.787×10^-2M

As for the second part, I seem to recall that AgCl is nearly insoluble in water, so if you add some Chloride ions into the left over solution then you would see a white precipitate if there are any silver ions still in solution. The Chloride ion can come from calcium chloride (clear colorless soln):

2 AgNO3 + CaCl2 -> 2 AgCl + Ca (NO3)2

Run it by a clever person in your lab section to check.

Pouring silver sheet ingot

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