Question

Questions and Problems

1. Complete the table containing the summary of data below.

Summary of Solubility and Ksp Data

Sat’d Ca(IO3)2 in: Solubility (M) _______ Ksp (M3 ) ___________T˚C ___________

H2O: Solubility (M) _______ Ksp (M3 ) ___________T˚C ___________

0.0100 M KIO3 : Solubility (M) _______ Ksp (M3 ) ___________T˚C ___________

0.0100 M KCl:  Solubility (M) _______ Ksp (M3 ) ___________T˚C ___________

2. Compare the values of Ksp of Ca(IO3)2 obtained for the three solvents.

(a) Which two are most nearly the same? Briefly explain. (

b) What value of Ksp would you expect for Ca(IO3)2 dissolved in 0.0100 M NaNO3 at the same temperature?

3. Is the solubility of Ca(IO3)2 greater in pure water or in 0.0100 M KCl? Briefly explain. {Hint: Consider the relative magnitudes of the ion-dipole and ion-ion forces experienced by the Ca 2+ and IO3 - ions in the two solutions.}

Chemical Equilibria: Ksp of Calcium Iodate Report Form -10

4. Calculate the molar solubility, s, of calcium iodate in 0.020 M Ca(NO3)2, a completely dissociated strong electrolyte. {NO3 - ion does not chemically interact with either Ca2+ or IO3 - .} Assume that Ksp for Ca(IO3)2 = 2.0 x 10-6 .

To set up the problem, we can write the following equations:

Material balance for calcium: [Ca2+] = s + 0.020

Material balance for iodate: [IO3 - ] = 2s

Ksp = 2.0 x 10-6 = [Ca2+][IO3 - ] 2 = (s + 0.020)(2s)2

Here in the last equation we have the solubility in s, but s is not negligible when compared to 0.020 M.

So here we are confronted with a cubic equation in s! To simplify the solution to this equation we can rearrange it into a more useful form and obtain s by successive approximations (iteration).

Solving the equation for s2 we get, s^2 = Ksp/ 4(s + 0.020) or s = 1 /2 [Ksp/( s + 0.020 )^]1/ 2

We have to find the value of s that makes both sides of the equation equal, that is the value of s we insert on the right side will yield the same value on the left side. We can start by neglecting s on the right side calculating s on the left side. The value of s obtained is then inserted on the right side and a new value of s is calculated on the left side. This iteration process is continued until two consecutive values of s calculated are the same within the precision of the measurement (significant figures).

Answer 1