A white crystalline solid compound A exhibits the following reactions: 1) The flame of a Bunsen burn — Physical Chemistry — Kinetics Chemistry Question
Chemistry of ions, stoichiometry, redox reactions
A white crystalline solid compound A exhibits the following reactions:
1) The flame of a Bunsen burner is intensively yellow coloured.
2) An aqueous solution of A is neutral. Dropwise addition of sulphurous acid (an SO2 solution) leads to a deep brown solution that is discoloured in the presence of excess of sulphurous acid.
3) If an AgNO3 solution is added to the discoloured solution obtained by 2) and acidified with HNO3, a yellow precipitate is obtained that is insoluble on addition of NH3, but can be readily dissolved by adding CN– or S2O3 2–.
4) If an aqueous solution of A is treated with KI and dilute H2SO4 a deep brown solution is formed that can be discoloured by addition of sulphurous acid or a Na2S2O3 solution.
5) An amount of 0.1000 g of A is dissolved in water, then 0.5 g KI and a few cm3 of dilute H2SO4 are added. The deep brown solution formed is titrated with 0.1000 M Na2S2O3 solution until the solution is completely discoloured. The consumption is 37.40 cm3.
What elements are contained in the compound A?
Model Answer
The solid must contain Na and I. The yellow colouration of the flame of the Bunsen burner indicates the presence of Na. A yellow silver salt that is dissolved only by strong complexing agents such as CN– or S2O3 2–, must be AgI.
What compounds can be considered as present on the basis of reactions 1) to 4)? Calculate their molar masses.
Model Answer
Reactions 1) to 4) indicate an Na salt of an oxygen containing acid of iodine:
Both SO2 and I− are oxidised. While in the first case I− is formed with an intermediate of I2 (or I3−, brown solution), in the second I2 (or I3−) is formed.
As the solution of A is neutral, NaIO3 and NaIO4 come into consideration.
M(NaIO3) = 22.99 + 126.905 + 3 × 16.000 = 197.895 = 197.90 g mol-1
M(NaIO4) = 22.99 + 126.905 + 4 × 16.000 = 213.895 = 213.90 g mol-1
Formulate the reactions corresponding to 2) to 4) for the compounds considered and write the corresponding equations in the ionic form.
Model Answer
2 IO3- + 4 H2O + 5 SO2 = 5 HSO4- + 3 H+ + I2
I2 + SO2 + 2 H2O = HSO4- + 3 H+ + 2 I-
IO4- + 7 I- + 8 H+ = 4 I2 + 4 H2O
IO3- + 5 I- + 6 H+ = 3 I2 + 3 H2O
I2 + 2 S2O3 2- = 2 I- + S4O6 2-
Decide on the basis of 5) which compound is present.
Model Answer
Experiment: 0.1000 g of the compound A ...... 3.740 × 10-3 moles S2O3 2-
1st hypothesis: The compound is NaIO3.
1 mole NaIO3 . . . . 197.90 g NaIO3 . . . . 6 moles S2O3 2-
0.1000 g NaIO3 . . . . 3.032 × 10-3 moles S2O3 2-
The hypothesis is false.
2nd hypothesis: The compound is NaIO4.
mole NaIO4 . . . . 213.90 g NaIO4 . . . . 8 moles S2O3 2-
0.1000 g NaIO4 . . . . 3.740 × 10-3 moles S2O3 2-
The compound A is NaIO4.