Answer to Question #153944 in General Chemistry for ray

Question

Formulas:

Cl2(g) + 2OH- (aq)--> ClO-(aq)+Cl-(aq)+H2O(l)

2H+(aq)+ClO-(aq)+2I-(aq)--> Cl-(aq) +I2(aq) +H2O(l)

I3-(aq)+2S2O32-(aq)--> 3I-(aq)+S4O62-(aq)

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

Given this information:

Molarity of Na2S2O3

0.125 M Initial Buret Reading

(Na2S2O3) 1.50 mL Final Buret Reading (Na2S2O3) 14.35 mL 

Question: Use the average volume and molarity of Na2S2O3 to determine the molarity of the diluted bleach. 

Answer 1

Answer:

The equation;

Cl_2(g) + 2OH⁻_(aq) → ClO⁻_(aq) + Cl⁻_(aq) + H₂O_(l)

Represents the chemical equation for the process of making the bleach by bubbling chlorine gas into sodium hydroxide solution.

The amount of hypochlorite ion present in a solution of the bleach is the one that is determined by an oxidation-reduction titration using iodine and thiosulfate ions.The iodide will be oxidized as thus; 2I⁻_(aq) →I_2(aq) + 2e⁻, due to its low reduction potential relative to other oxidizing agents.In acidic solution, hypochlorite ions oxidize iodide ions to form iodine, I₂.The iodine that forms is then titrated with a standard solution of sodium thiosulfate.The analysis takes place in a series of steps represented by the three equations:

STEP I: 2H+(aq) + ClO−(aq) + 2I−(aq) →Cl−(aq) + I2(aq) + H2O (l)

STEP II: I2(aq) + I−(aq) →I3−(aq)

STEP II : I3−(aq) + 2S2O32−(aq) →3I−(aq) + S4O62−(aq)

Calculations;

Average volume of N2S2O3

= 14.35mL - 1.50mL

= 12.85 mL

Moles of Na2S2O3 used;

since 0.125 mol $\implies$ 1000mL

? mol $\implies$ 12.85mL

= $\dfrac{0.123molx 12.85mL}{1000mL}$ = 1.581 x 10^-3 mol

From the balanced equation in step III above, mole ratio of thiosulphate: triiodide ions = 2 : 1

Thus moles of Triiodide ions = $\dfrac{1}{2}$ x 1.581 x 10^-3 mol

= 7.903 x 10^-4mol

From the balanced equation in step II, mole ratio of I₃^- : I₂^- is 1:1

Thus moles of Iodide molecule, I₂^- is 7.903 x 10^-4 mol

Also from equation in step I mole ratio of I2 : ClO^- is also 1:1, hence moles of ClO^- is 7.903 x 10^-4, which is also the number of moles of Cl^- in the diluted bleach.

Molarity of diluted bleach is calculated as thus;

7.903 x 10^-4 moles $\implies$ 25mL (pipette)

? moles $\implies$ 1000mL

$= \dfrac{7.903x10-^4 x 1000mL}{25mL} = 0.0316M$