Homework Help: Chem- Please check :)

Posted by Anonymous on Wednesday, April 18, 2012 at 10:21pm.

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, Wednesday, April 18, 2012 at 10:26pm
Except for the number of significant figures, yes.
The 0.0129g limits you to three in the answer.
Chem- Please check :) - Anonymous, Wednesday, April 18, 2012 at 10:30pm
I don't get it :S
Chem- Please check :) - DrBob222, Wednesday, April 18, 2012 at 10:54pm
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 1.89E-12.
Here is a link if you want to read more about s.f.
http://www.chemteam.info/SigFigs/SigFigRules.html
Chem- Please check :) - Anonymous, Wednesday, April 18, 2012 at 11:33pm
oh okay. make sense :) thank you so much!

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