Answer :
The solubility (in mol/L) of silver chromate in 1.20 M in potassium chromate, silver nitrate and in pure water is
- X = 5.0*10^-7 M
- X'' = 8.333*10^-13 M
- X'''= 6.694*10^-5 M
This is further explained below
What is the solubility (in mol/L) of silver chromate in 1.20 M potassium chromate aqueous solution?
Generally,The Chemical reaction is
Ag2CrO4 <----> 2 Ag+ + CrO42-
Ksp = [Ag+]^2[CrO42-]
1.12×10−12 = [Ag+]^2[CrO42-]
Hence
1.12×10−12.=(2X)^2*(1.2)
X = 5.0*10^-7 M
b)
1.12×10−12=(1.2)^2*(X)
1.12×10−12= 1.44 * 1(X)^1
X'' = 8.333*10^-13 M
In conclusion
1.12×10−12=(2X)^2*(s)
1.12×10−12= 4(X)^3
X'''= 6.694*10^-5 M
Read more about Chemical reaction
https://brainly.com/question/16416932