When calcium iodate is added to water at 25 degrees Celsius, the equilibrium concentration of calcium ion was found to be 0.011M. Calculate the value of Ksp for this salt.
.......Ca(IO3)2 ==> Ca^2+ + 2IO3^-
I......solid.........0.......0
C......solid.........x.......2x
E......solid.........x.......2x
Ksp = (Ca^2+)(IO3^-)^2
Ksp = (x)(2x)^2 = 4x^3 and the problem tells you x is 0.11M. Substitute and solve for Ksp.
Thank you!
To calculate the value of Ksp (solubility product constant) for calcium iodate (Ca(IO3)2), we need to use the given equilibrium concentration of calcium ions.
The balanced equation for the dissolution of calcium iodate in water is:
Ca(IO3)2 (s) ⇌ Ca2+ (aq) + 2IO3- (aq)
The equilibrium expression for the solubility of calcium iodate is:
Ksp = [Ca2+][IO3-]^2
Given that the equilibrium concentration of calcium ion ([Ca2+]) is 0.011 M, we can substitute this value into the Ksp expression:
Ksp = (0.011)(2 * 0.011)^2
Simplifying the expression:
Ksp = 0.011 * 0.044
Ksp = 0.000484
Therefore, the value of Ksp for calcium iodate is 0.000484.