Calculate the molar solubility, s, of calcium iodate in 0.020 M Ca(NO3)2, a comp

Question

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 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

The last equation only contains one unknown, the value of s, which we would like to calculate; but s is not negligible compared to 0.020, which leaves us with a nasty equation that is cubic in s. An approach that often works in a situation like this is to rearrange the equation into a more useful form and obtain s by iteration. Thus we can write

4s^2 = Ksp/(s + 0.020) or s = 1/2(Ksp/(s + 0.020))^1/2

We must find the value of s that makes both sides of the equation equal. Start first by inserting a trial value of s (say, 0.006M) in the right hand side of the latter equation and calculating a value of s. You will get s = 0.00438. For your next trial vvalue. take something about halfway between 0.00438 and 0.006. Continue the iteration process until you calculate a value of s that is nearly the same (within 5%) as the value you inserted on the right-hand side of the equation. Thank you!)

Explanation / Answer

do not worry it is simple problem . please follow step by step .

aim: we need to calculate molar solubility of calcium iodate in 0.020 M Ca(NO3)2

calcium iodate formula = Ca(IO3)2 and it is sparingly soluble salt.

Ca(IO3)2 ksp is given = 2.0 x 10^-6.

Ca(IO3)2 -------------------------------> Ca+2 + 2IO3-

                                                        S               2S

S here solubility

Ca(NO3)2 strong electrolyte completely dissociate

Ca(NO3)2 -------------------------> Ca+2 + 2 NO3 -

0.020 M                                      0.020      2 x 0.020

now in the solution Ca+2 ion concentration = S + 0.020

here 'S' from Ca(IO3)2 and   0.020 from Ca(NO3)2

now in presence of common ion Ca+2 equilibrium

Ca(IO3)2 <-------------------------------> Ca+2 + 2IO3-

                                                      S+ 0.020            2S

                                                      0.020              2S

(compare to 0.020 we neglet S for Ca+2 )

Ksp = [Ca+2][ IO3-]^2

2.0 x 10^-6. = (0.020) (2S)^2

S^2 = 2.5 x 10^-5

S= 5 x 10^-3

S= 0.005M -----------------------------------> (answer)

molar solubilty of calcium iodate = S = 0.005M    in presence of common ion solution calcium nitrite with 0.020M

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