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solubility

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ADG:
Here's the question I got on one of my tests that I just cannot figure out.  It was only worth 2 marks but it's been bugging me...  It's probably just common sense but could some one help me out and care to explain this one for me?

In attempting to prepare a 0.1 M solution of silver nitrate in tap water, a student claimed that she could not dissolve the silver nitrate because the solution remained couldy.  Explain the student's problem.  The solubility of AgNO3 is 245g in 100g water at 25 degrees C.

mike:
Silver nitrate will form insoluble silver chloride with Cl- ions in solution. More than likely tap water will have some chloride ions in it.

lemonoman:
That'd be it.

It said that the solubility = 245g AgNO3 / 100 mL
245 g AgNO3 / (169.873 g/mol) = 1.442 mol AgNO3

So solubility = 1.442 mol / 100 mL = 14.42 mol/L = 14.42 M

0.1 M AgNO3, therefore, should be REALLY easy to make.  This is what tells us it's something OTHER than insolubility of AgNO3

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