If I Had 50.00mL of Saturated Silver Sulfate Solution, When Evaporated Leaving 0.25g of Silver Sulfate What Is?
Posted by junkgurl in Junk Silver Coin FAQ, tags: 0.25g, 50.00mL, evaporated, leaving, saturated, silver, solution, sulfateQuestion by scienceguy: If I had 50.00mL of saturated silver sulfate solution, when evaporated leaving 0.25g of silver sulfate what is?
If I had 50.00mL of saturated silver sulfate solution, when evaporated leaving 0.25g of silver sulfate what is the
a) Amount of silver sulfate in the 50ml sample: .25g?
c)molar concentration of silver ion in saturated solution
d) molar concentration of sulfate ion in saturated solution
e) what is ksp expression of silver sulfate and how do i calculate it
Thanks for all the help guys 
Best answer:
Answer by Trevor H
a) correct - 0.25g This does not agree with my solubility table for solubiity of Ag2SO4 My list says a saturated solution will contain 0.6g Ag2SO4 per 50ml saturated solution. I will use your figures.
Molar mass Ag2SO4 =311.799 g/mol
b) You have 0.25g in 50ml. How much in 1000ml? 1000/50*0.25 =5g
This is 5.0/311.799 = 0.0160mol per litre. Which is the molar solubility.
c) Ag2 = 107.87*2 = 215.74.
Molar conc. of Ag ions = 215.74/311.799*0.0160 = 0.011mol/litre
d)Molar conc. of SO4 ions = 92.066/311.799*0.016 = 0.00493 mol/litre
e) Ag2SO4 (s) → 2 Ag+ (aq) + SO4 2¯ (aq)
Ksp = [Ag+]² * [SO4 2¯ ]
Ksp = (0.011)² * (0.00493)
Ksp = 5.965 * 10^-7
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Originally posted 2010-08-30 13:43:14. Republished by Blog Post Promoter
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