Tuesday
August 4, 2015

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? ^

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Members