In: Chemistry

Questions and Problems 1. Complete the table containing the summary of data below. Summary of Solubility...

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).

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