1) calculate the molar solubility of AgCl at 25C in: a) pure water and b) 2.0 M
ID: 918175 • Letter: 1
Question
1) calculate the molar solubility of AgCl at 25C in: a) pure water and b) 2.0 M NH3 2) calculate the solubility in g/L of AgBr in a) pure water and 0.01 M NaBr. 3) if 2.5 g of copper sulfate is dissolved in 900mL of 0.3M NH3 what are the concentrations of Cu^2+ , Cu(NH3)4^2+ and NH3 at equilibrium. Please help :) 1) calculate the molar solubility of AgCl at 25C in: a) pure water and b) 2.0 M NH3 2) calculate the solubility in g/L of AgBr in a) pure water and 0.01 M NaBr. 3) if 2.5 g of copper sulfate is dissolved in 900mL of 0.3M NH3 what are the concentrations of Cu^2+ , Cu(NH3)4^2+ and NH3 at equilibrium. Please help :) 2) calculate the solubility in g/L of AgBr in a) pure water and 0.01 M NaBr. 3) if 2.5 g of copper sulfate is dissolved in 900mL of 0.3M NH3 what are the concentrations of Cu^2+ , Cu(NH3)4^2+ and NH3 at equilibrium. Please help :)Explanation / Answer
1a) Ksp= 1.8x10-10
AgCl <-------> Ag+ + Cl-
----------- 0 0
----------- x x
Ksp=[Ag+][Cl-]= x2 ---> x= 1.34x10-5
1b) KfAg(NH3)2+=1.6x107
AgCl <-------> Ag+ + Cl-
Ag+ + 2NH3 ------> Ag(NH3)2+
-------------------------------------------------------
AgCl + 2NH3 -----> Cl- + Ag(NH3)2+
K=[Ag(NH3)2+][Cl-]/[NH3]2= KfAg(NH3)2+ x Ksp=2.88x10-3
The NH3 is being consumed so it will be -x but we have 2 moles so it will be -2x. The starting concentration is 2M so we have 2-2x. Cl- and Ag(NH3)2+ are being created so they will be x.
K= x2/(2-2x)2 ----> x= 0.09693
2) Ksp AgBr=5.4x10-13= [Ag+][Br-]= x2 ----> x= 7.35x10-7
In NaBr:
Ksp=(x)(x+0.01)=5.4x10-13 ----> x= 5.4x10-11
3) I can´t figure out the last one, really sorry :(
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