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Chemical equilibrium


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Anyone here did chemical equilibrium? I'm stuck on this question. Any help would be much appreciated.

 

1) for AgCl has a value of 1.8 x 10-10 at 25 degrees celsius, calculate whether or not 1 mg AgCl could dissolve in 1.0 L water.If Ksp

 

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

Ksp = [Ag+] [Cl-]

= [x] [x]

= [x2]

x2= 1.8 x 10^-8

[x] = 1.34x10^-5 mol/l

 

= 143g/mol

= 0.0018g/L ??????

 

2) Although the solubility of AgCl is extremely low in pure water, it is even lower in a 0.1M solution of sodium chloride (NaCl) due to the ‘common- ion effect’. Briefly explain this observation.

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Common Ion Effect results from LeChatelier's Principle. Basically, if we mix the soluble salt with the slightly soluble salt together we increase the concentration of the common ion, in this case, Cl-. This will place a, for lack of a better term, "stress" upon the equilibrium of the AgCl, the slightly soluble salt, due to the added concentration. The equilibrium will react trying to undo the stress applied due to the added common ion, thus causing it to shift to the left. This reduces the slightly soluble salt's solubility.

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Taking more formal approach:

 

Kso = [Ag+][Cl-]

 

If there is nothing else in the solution, [Ag+] = [Cl-] and that's the case you already did.

 

In the presence of Cl-:

 

[Ag+] = Kso/[Cl-]

 

and the [Ag+] is the amount of AgCl dissolved.

 

Best,

Borek

--

Chemical calculators for labs and education

http://www.chembuddy.com

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