this is the experiment of : The Solubility roduct Constant of Calcium Iodate, Ca
ID: 520481 • Letter: T
Question
this is the experiment of :
The Solubility roduct Constant of Calcium Iodate, Ca(IO3)2
can you please help me with the calaclations for the lap report . thanks so much
Theory
A. The molar solubility and the solubility product constant of Ca(IO3)2
Calcium iodate, a salt whose solubility you will study in this experiment, has been shown to completely dissociate in water into calcium, Ca2+, and iodate, IO3- ions.
Ca(IO3)2(s) « Ca2+(aq) + 2IO3-(aq) (1)
The equilibrium constant expression for this reaction (often called the solubility product expression) is written as shown in Equation (2), observing the convention that the activity of the pure solid is taken to be equal to 1.00 so that it does not appear in the equilibrium constant expression.
Ksp = [Ca2+][IO3-]2 (2)
If we let the symbol s represent the molar solubility of calcium iodate (in mol/L), the concentration of Ca2+ and IO3- ions will be related to the molar solubility, s, by the following equations:
[Ca2+] = s (3)
[IO3-] = 2s (4)
By substituting this into Equation (2),
Ksp = [Ca2+][IO3-] 2= s(2s)2 = 4s3 (5)
B. The solubility of Ca(IO3)2 in KIO3 Solution: The common ion effect
From Le Chatelier’s principle we would predict that the molar solubility of calcium iodate would be smaller in a solution of potassium iodate, KIO3, which is a strong electrolyte that completely dissociate in water. The addition of KIO3 would shift the equilibrium shown in Equation (1) toward the left, decreasing the amount of calcium iodate that dissolves. Such a decrease in solubility that occurs when a salt is dissolving in a solution that already contains one of the salt’s ions is called the common ion effect.
To test this hypothesis you will measure the solubility of calcium iodate in 0.01 M KIO3. Under these conditions the concentrations of the ions will be related to the molar solubility of Ca(IO3)2 in the following way.
[Ca2+] = s (6)
[IO3-] = 0.01 + 2s (7)
Note that all of the calcium ion must come from dissolved calcium iodate. However, the iodate ion comes from both KIO3 and dissolved Ca(IO3)2. Now, the solubility product will be given by,
Ksp = [Ca2+][IO3-]2 = s(0.01 + 2s)2 (8)
If we know the concentration of KIO3, a single measurement of the total iodate concentration allows us to calculate the molar solubility of calcium iodate and the solubility product constant.
Determination of the Molar Solubility of Calcium Iodate
The molar solubility of Ca(IO3)2 can be determined by measuring either the concentration Ca2+ or the concentration IO3-. We will use a procedure for measuring the IO3- concentration that makes use the fact that IO3- oxidizes iodide ion.
IO3- + 5I- + 6H+ ® 3I2 + 3H2O (9)
The I2 produced is in turn titrated with Na2S2O3.
I2 + 2S2O32- ® 2I- + S4O62- (10)
Note from the overall stoichiometry that each mole of IO3- will produce enough I2 to consume 6 moles of S2O32-. Therefore, the concentration of IO3- ion will be given by
[IO3-] = (volume of S2O32- used) x (molarity of S2O32- ) (11)
(volume of IO3- used) x 6
Procedure
A. The Molar Solubility of Ca(IO3)2 in Pure Water
Fill a 50 mL buret with 0.05 M Na2S2O3 solution.
Obtain about 30 mL of saturated Ca(IO3)3 solution in a clean dry flask.
Put 50 mL DI water and 10mL of 1M KI into a 250 mL Erlenmeyer flask, and swirl.
Transfer 10.0 mL of saturated Ca(IO3)2 solution and 10 mL of 1.0 M HCl into Erlenmeyer flask. (Now the solution will turn brown)
Immediately titrate the solution with 0.05 M Na2S2O3 until the solution is yellow.
Then add 5 mL of 0.1% starch indicator. The solution will turn blue-black.
Continue the titration until you get a sharp change from blue to a colorless solution.
Repeat the titration procedure with a second sample.
B. The Molar Solubility of Ca(IO3)2 in 0.01 M KIO3
Use the procedure described in Part A to titrate two 10.0 mL samples of the saturated solution of Ca(IO3)2 in 0.01 M KIO3.
Calculations A. The Molar Solubility of Ca0O302 in Pure Water 1. Calculate the concentration of IO3-ion in the saturated solution (use Equation 11). 2. Find the molar solubility of Ca TO3)2 see Equations 3 and 4) 3. Calculate the solubility product constant for CaIO3)2 use equation 5 B. The Molar Solubility of Ca O302 in 0.01 MKIO3 1. Calculate the total IO3-ion concentration as in Part A. 2. Subtract 0.01Mfrom the total concentration to obtain the concentration of IO3 ions that come from dissolved Ca(IO3)2. 3 Divide this result by two to obtainthemolar solubility of Ca IO32. Is it smaller as we predicted byLe Chatliers principle 4. Calculate the solubility product constant for calciumiodate (see Equation 8) Does the calculatedvalue of Ksp agreewith the value of Ksp calculated in Part A.?Explanation / Answer
In this iodometric titration, the iodate ions react with iodide ions and acidic protons to liberate iodine. This liberated iodine is then titrated with sodium thiosulfate. The balanced reactions are given as:
IO3(aq) + 5I(aq) + 6H+(aq) -----> 3I2(aq) + 3H2O(l)
I2(aq) + 2S2O3-2(aq) ------> 2I(aq) + S4O6-2(aq)
From the above equation we know that 1 mol of I2 is consumed per 2 mol of S2O3-2
And 3 mol of I2 are formed per mol of IO3
Therefore, 6 mol of S2O3-2 are consumed per mol of IO3
Number of moles of IO3 required = 6 X mol of S2O3-2
Number of molesof S2O3-2 = concentration of S2O3-2 solution X volume of S2O3-2 solution
Number of moles of IO3 required = 6 X concentration of S2O3-2 solution X volume of S2O3-2 solution
For Part A:
Number of moles of IO3 required = 6 X 0.05 mol L-1 X 16.7 X 10-3 L
NB: in titrations you use two concurrent readings. Group 3 readings are quite close to each other and should be used. Since your calculations are based on Group 1 readings, I have proceeded with the same by averaging the two readings. (.16.7 mL).
Number of moles of IO3 required = 5.01 X 10-3 mol
[IO3] = Number of moles of IO3 required / Volume of IO3 pipetted out
= 5.01 X 10-3 mol / 30 X 10-3 L
= 0.167 mol L-1
Ca(IO3)2 (s) <=> Ca2+(aq) + 2(IO3)-(aq)
From the above equation, Ksp = [Ca2+]e [(IO3)-] 2e
If s is the solubility of [Ca2+] at equilibrium then the solubility of [(IO3)-]will be 2s
[Ca2+] = ½ [IO3]
= 0.167 / 2 mol L-1 = 0.0835 mol L-1
Ksp = 0.0835 (0.167)2
= 0.0835 X 0.02789
=2.33 X 10-3
Note, Ksp does not have units.
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