16.1 mL of 0.0478 M sodium thiosulfate is used to reach an end point with 10 mL

Question

16.1 mL of 0.0478 M sodium thiosulfate is used to reach an end point with 10 mL of Calcium Iodate. Using math we find that 7.696x10-4 mols of thiosulfate was used in the rxn. Since there is a 1 to 6 ratio of Ca(IO3)2 to thiosulfate, this means 1.283x10-4 mols of iodate was present in the the rxn. 1.283x10-4 mols/10ml = 0.0128 M for the Ca(IO3)2.

Now I set up the icetable how i think i need to but correct me if I'm wrong:

dissassiocation of the Ca(IO3)2

Now, using everything i've given, how do I

1.determine the solubillity of the Ca(IO3)2

2. Caluclate the Ksp of the Ca(IO3)2

Explanation / Answer

The titration reaction-

IO3-+5I- +6H+--->3I2 +3H2O

2S2O32- +I2 --->S4O62-+2I-]*3

---------------------------------------------------------------------------

net rxn-   IO3-+6S2O32-++6H+--->3S4O62-+3H2O+I-

For Ca(IO3)2---->Ca2+ +2IO3-

the titration rxn would be-Ca(IO3)2+12S2O32-++12H+--->6S4O62-+6H2O+2I- +Ca2+ +24Na+

Calcium iodate and sodium thiosulfate react in 1:12 molar ratio

moles of sodium thiosulfate reacted=0.0478mol/L*16.1ml*10^-3L/ml=7.696*10^-4 moles

moles of calcium iodate reacted=1/12 *7.696*10^-4 moles=6.413*10^-5 moles

molarity of calcium iodate=6.413*10^-5 moles/10ml=6.413*10^-5 moles/0.01L=6.413*10^-3 M

you have found out the actual moles of iodate present in the solution after dissolution=solubility of iodate=[IO3-]eq=6.413*10^-3 M

Let the solubility of iodate be S,then

Ca(IO3)2---->Ca2+ +2IO3- gives

S=[IO3-]=6.413*10^-3 M

[Ca2+]=1/2*S=1/2*6.413*10^-3 M=3.206*10^-3M

ksp=[Ca2+] [IO3-]=(3.206*10^-3 M)(6.413*10^-3 M)=2.056*10^-5

ksp=2.056*10^-5

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