Posted by Anonymous on .
The solubility of silver chromate is 500 ml of water at 25 degrees celsius is 0.0129g. Calculate its solubility product constant.
n = m/M
= 0.0129g/331.73g/mol
= 3.888 x 10^5mol
concentration (ag2crO4)=n/V
=3.888x10^5/0.5L
= 7.777 x 10^5
ag2crO4 <> 2ag+ + CrO4
ksp= [ag+]2 x [CrO4]
= (2x)^2 x (x)
= 4(7.777 x 10^5)^3
= 1.888 x 10^12
is that right? ^

Chem Please check :) 
DrBob222,
Except for the number of significant figures, yes.
The 0.0129g limits you to three in the answer. 
Chem Please check :) 
Anonymous,
I don't get it :S

Chem Please check :) 
DrBob222,
In multiplication and division you are allowed as many significant figures in the answer as you have in the least value of the numbers multiplied or divided. You have three significant figures in the 0.0129 grams and 5 in the molar mass; therefore, you are allowed three in the answer. Your answer should be rounded to 1.89E12.
Here is a link if you want to read more about s.f.
http://www.chemteam.info/SigFigs/SigFigRules.html 
Chem Please check :) 
Anonymous,
oh okay. make sense :) thank you so much!