Could someone please advise if I have answered the question correctly?
Thanks!
Calculate the Ksp of Ag2CrO4. ..
Ksp measured under normal conditions is 25 degrees C
Ag2CrO4 ↔ 2Ag1+ + CrO4
Ksp = [Ag1+]2 + [CrO42-]
ICE Molar Solubility [Ag1+] in mol/L [CrO42-] in mol/L
Initial x
Change +2x +x
Equilibrium 2x x
1.1 X 10-12 = [2x]2
1.1X 10-12 =4x3
1.1X10-12 /4 = x3
2.75X10-13= x3
3√2.75X10-13=x
6.5X10-5=x
The molar solubility of Ag2CrO4 is 6.5X10-5