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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 :) -

Except for the number of significant figures, yes.
The 0.0129g limits you to three in the answer.

• Chem- Please check :) -

I don't get it :S

• Chem- Please check :) -

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.
http://www.chemteam.info/SigFigs/SigFigRules.html

• Chem- Please check :) -

oh okay. make sense :) thank you so much!