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? ^

Except for the number of significant figures, yes.

The 0.0129g limits you to three in the answer.

I don't get it :S

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

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

Yes, your calculation is correct. The solubility product constant (Ksp) for silver chromate (Ag2CrO4) is 1.888 x 10^-12.

To calculate the solubility product constant, you first need to determine the concentration of the dissolved ions in the solution.

Given that the solubility of silver chromate is 0.0129g in 500 ml of water, you can calculate the number of moles (n) using the formula n = m/M, where m is the mass and M is the molar mass of silver chromate.

In this case, n = 0.0129g / 331.73g/mol = 3.888 x 10^-5 mol.

Next, you calculate the concentration of silver chromate (Ag2CrO4) by dividing the number of moles (n) by the volume (V) of the solution. In this case, the volume is given as 500 ml, which is equivalent to 0.5 L.

So, the concentration of Ag2CrO4 = n / V = 3.888 x 10^-5 mol / 0.5 L = 7.777 x 10^-5 M.

Finally, you use the balanced equation for the dissolution of Ag2CrO4 to determine the solubility product constant. In this case, Ag2CrO4 dissociates into 2Ag+ ions and 1CrO4- ion.

The Ksp expression is Ksp = [Ag+]^2 x [CrO4-]. Since the concentration of Ag+ and CrO4- ions is the same (x), you can express the Ksp as:

Ksp = (2x)^2 x (x)

Substituting the concentration of Ag+ and CrO4- as 7.777 x 10^-5 M, you get:

Ksp = 4(7.777 x 10^-5)^3 = 1.888 x 10^-12.

So, the solubility product constant for silver chromate is 1.888 x 10^-12.