what are the procedures and materials needed to make a 0.10m solution of Silver(I) Chloride?
3 answers
What course is this? How advanced? I don't think you can put 0.1 mol AgCl in solution without using some complexing agent. And then it isn't AgCl.
This is an Advance placement chemistry class. We were told to fine the materials needed to do this lab, and the procedure. i need this info so i can physically do the lab.
Without knowing the details of the experiment I'm at a disadvantage to know exactly how to answer. However, AgCl is insoluble in water, the Ksp = 1.82E-10 or about 1.35E-5 moles/L which is equivalent to about 0.0019 g/L. So you can see that AgCl, without some help from the outside, is not that soluble. I suggest two approaches.
a. If the AgCl solution is to be JUST AgCl and the solvent, you might look into the solubility of AgCl at temperatures higher than 25C.
b. If something can be added to increase the solubility, then add NH3(aq). That forms the Ag(NO3)2^+2 complex ion and dissolves AgCl completely (if there is enough NH3 present.
So you need 0.1 moleAgCl/1000 g H2O/NH3.
I would prepare such a solution by weighing out 0.1 mol AgCl (about 0.1 x 143 = ?? grams), adding aq NH3 drop by drop to dissolve it, then add 1 kg water. I don't know how accurately this is to be done but that way of doing it will add MORE than 1 kg solvent. If better accuracy is needed, you could add the NH3 solution, drop by drop from a buret, calculate the mass NH3 soln from the density, then add the difference of water to make 1 kg. Another way is to weigh the NH3 soln, add drop by drop to the AgCl until all is dissolved, reweigh the NH3 soln to determine how much of that was added, then make up the difference with water to make 1 kg. All of this is shooting in the dark because I don't know what you are to do with the AgCl solution once prepared.
a. If the AgCl solution is to be JUST AgCl and the solvent, you might look into the solubility of AgCl at temperatures higher than 25C.
b. If something can be added to increase the solubility, then add NH3(aq). That forms the Ag(NO3)2^+2 complex ion and dissolves AgCl completely (if there is enough NH3 present.
So you need 0.1 moleAgCl/1000 g H2O/NH3.
I would prepare such a solution by weighing out 0.1 mol AgCl (about 0.1 x 143 = ?? grams), adding aq NH3 drop by drop to dissolve it, then add 1 kg water. I don't know how accurately this is to be done but that way of doing it will add MORE than 1 kg solvent. If better accuracy is needed, you could add the NH3 solution, drop by drop from a buret, calculate the mass NH3 soln from the density, then add the difference of water to make 1 kg. Another way is to weigh the NH3 soln, add drop by drop to the AgCl until all is dissolved, reweigh the NH3 soln to determine how much of that was added, then make up the difference with water to make 1 kg. All of this is shooting in the dark because I don't know what you are to do with the AgCl solution once prepared.
1 answer