a 2 ml sample was taken of a saturated solution of barium iodate at 10 degrees C

Question

a 2 ml sample was taken of a saturated solution of barium iodate at 10 degrees Celsius. The sample was mixed with 2 ml of 0.5 M KI and 2 ml of 1 M HCl. the transmittance of the mixture was read and compared to a calibration curve which indicated a concentration of I3- as 5.52*10-4 M calculate the following.
a. the moles of I3- in the mixture.
b. the moles of IO3- in the original sample.
c. the concentration in moles/L of IO3- in the original sample.
d. the concentration of Ba2+ (in M) in the original sample.
e.the Ksp of barium iodate.

Explanation / Answer

The reactions taking place are:

Ba(IO3)2 (s) <=====> Ba2+ (aq) + 2 IO3- (aq) ……(1)

IO3- (aq) + 5 I- (aq) + 6 H+ (aq) -------> 3I2 (aq) + H2O (l) …….(2)

I2 (aq) + I- (aq) ------> I3- (aq) ……(3)

a) Molar concentration of I3- was determined to be 5.52*10-4 M = 5.52*10-4 mol/L.

Volume of the solution = (2.0 + 2.0 + 2.0) mL = 6.0 mL = (6.0 mL)*(1 L/1000 mL) = 0.006 L.

Moles of I3- = (volume of solution in L)*(concentration of I3- in mol/L) = (0.006 L)*(5.52*10-4 mol/L) = 3.312*10-6 mole (ans).

b) As per stoichiometric equation (3),

1 mole I3- = 1 mole I2

Again, as per equation (2),

1 mole IO3- = 3 moles I2

Combining equations (2) and (3), we have

3 moles I3- = 1 mole IO3-

Therefore, 3.312*10-6 mole I3- = (3.312*10-6 mole I3-)*(1 mole IO3-/3 moles I3-) = 1.104*10-6 mole IO3- (ans).

c) Concentration in mol/L of IO3- in the original sample = (moles of IO3-)/(volume of IO3- taken in L) = (1.104*10-6 mole)/[(2.0 mL)*(1 L/1000 mL)] = 5.52*10-4 mol/L = 5.52*10-4 M (ans).

d) As per stoichiometric equation (1),

1 mole Ba2+ = 2 moles IO3-.

Therefore, moles of Ba2+ = (1.104*10-6 mole IO3-)*(1 mole Ba2+/ 2 moles IO3-) = 5.52*10-7 mole.

Concentration of Ba2+ = (5.52*10-7 mole)/(0.002 L) = 2.76*10-4 mol/L = 2.76*10-4 M.

e) Ksp = [Ba2+][IO3-]2 = (2.76*10-4)*(5.52*10-4)2 = 8.41*10-11 (ans).

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