1. Even though The solubilities of calcium iodate change in pure water and in 0.
ID: 906449 • Letter: 1
Question
1. Even though The solubilities of calcium iodate change in pure water and in 0.0100M potassium Iodate, the Ksp values will stay the same. Explain why this is so2. Explain using Le Chatelier's principal why the molar solubility decreases when Ca(IO3)2 was in a 0.0100M KIO3 solution Concentration of Na.so, solution .M Part 1 Initial biuet reading (to the nearest 0.1 ml) Final buret reading (to the nearest O.1 ml) Volune of Na.S.O, used -C Trial 3 Temperature of Sanrahon of Cados). Solitions CalO,), in Pure Water Trial 1 Trial 2 ml Part 2Calo In 0.0100MM KI0, Calo) in 00100M KIO Initial buret reading (to the nearest 0.1 ml) Final buret reading (to the nearest 0.1 ml Volune of Na.S,0, used ml Calculations Part 1 CalO,, in Pure Water Trial 1 Trial 2 Trial 3 Molarity of the 1O, (fromm equation 6) Average Molarity of the 10, Molarity of the Ca Value of Kip for Cato(equation (3) on page D) (1/2 the above value) 292x/D 6 Trial 2 Trial 3 Part 2Ca(10y) in 0.0100M KIO Molanty of the I0, Trial 1 (from equation 6) olsers Average Molarity of the IO Molarity of the Ca Value of Kip for Ca(o) equation (3) on page l) (1/2 the (above value 00100) 12. Compare [Ca-+] from Parts l and 2 (note: this is the solubility. Is Le Chatelier's law followed? Explain Arethee'sfron,Parts land 2consister des they ate consise Are the Kips from Parts 1 and 2 consisteat? es rey le Con
Explanation / Answer
1.
Ksp is a constant. its value is independent of concentration and hence from solubility. Ksp can be changed only by changing temperature.
This is the reason change is solubility didn't chnage the Ksp, since temperature was not changing
2.
Ca(IO3)2 ---> Ca2+ + 2IO3-
So when you add KIO3 , you are adding more of IO3-, which according to Le Chatelier's principal will shift the equilibrium towards left. hence more Ca(IO3)2 will be formed and so solubility is said to decrease
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