Can anyone help me?
Ksp of CuCO3 = 3.5x10^ -4
Ksp of Ca(OH)2 = 8.9x10^ -5
Ksp of Ag2CrO4 = 10.0x10^-11.
Given by these, Which has the highest molar solubility? Thanks
Determine the solubility of each using Ksp then compare the numbers. Pick the largest number. Here is how you do Ag2CrO4.
..............Ag2CrO4 ==> 2Ag^+ + CrO4^2-
I..............solid.................0...............0
C.............solid................2x..............x
E..............solid...............2x...............x
Ksp = (Ag^+)^2(CrO4)
Substitute the E line into Ksp expression and solve for x = solubility of Ag2CrO4 in moles/L.
Ksp = (2x)^2 (x)
Thank youuuu. :D
To determine which compound has the highest molar solubility, we need to compare their solubility product constants (Ksp) values. The compound with the highest Ksp will have the highest molar solubility.
The molar solubility of a compound represents the concentration of an ion in a saturated solution. The higher the concentration of ions (from the dissociation of the compound) in the solution, the higher the molar solubility.
Let's compare the Ksp values of the given compounds:
1. CuCO3: Ksp = 3.5x10^-4
2. Ca(OH)2: Ksp = 8.9x10^-5
3. Ag2CrO4: Ksp = 10.0x10^-11
Based on their Ksp values, we can see that Ag2CrO4 has the smallest Ksp value (10.0x10^-11), indicating the lowest molar solubility among the three compounds.
Comparing CuCO3 and Ca(OH)2, CuCO3 has the higher Ksp value (3.5x10^-4) compared to Ca(OH)2 (8.9x10^-5). Therefore, CuCO3 has the highest molar solubility among the three compounds.
In conclusion, CuCO3 has the highest molar solubility among CuCO3, Ca(OH)2, and Ag2CrO4, based on their respective Ksp values.