Question

mass of Na2S2O3 is 1.220g

titration of bleach with Na2S2O3 solution:

trial 1: total volume used 66mL

trial 2: total volume used 73.8mL

Density of Bleach 1.074 kg/L

a) find the moles of Na2S2O3 used to titrate the bleach solution

b)find the moles of I2 formed by reaction with bleach

c) find the molarity of OCl- present in the bleach solution

d) find the grams of NaOCl present in the bleach

e) find the percent NaOCl present in the bleach

Answer 1

(a) We know

Number of moles = given mass in g / gram molar mass of substance

given mass of Na₂S₂O₃ = 1.220 g

gram molar mass of Na₂S₂O₃ = 158 g/mol

therefore

Number of moles of Na₂S₂O₃ = 1.220g / 158 g/mol

= 0.0077 mol

Hence Number of moles of Na₂S₂O₃ = 0.0077 mol

(b) Reaction of I₂ with Na₂S₂O₃ is:

I₂ + 2 S₂O₃^2- -----------------> S₄O₆^2- + 2 I^-

from this balanced chemical equation it is cleared that

2 mol of Na₂S₂O₃ reacts with = 1 mol of I₂

therefore

0.0077 mol of Na₂S₂O₃ reacts with = 1/2 x 0.0077 mol of I₂

                                                         = 0.00385 mol of I₂

Hence number of moles of I₂ formed = 0.00385 mol

(c) Formation of I₂ by reaction of I^- with NaOCl is :

OCl^-(aq) + 2 I^-(aq) + 2H₃O⁺ (aq) ------------> I₂ (aq) + Cl^- (aq) + 3H₂O (l)

from balanced chemical equation

1 mol of I₂ formed by = 1 mol of OCl^-

therefore

0.00385 mol of I₂ formed by = 0.00385 mol of OCl^-

so number of moles of OCl^- = 0.00385 mol

also average volume = 73.8 mL + 66 mL / 2 = 139.8 / 2   = 69.9 mL = 0.0699 L

we know Molartity = Number of moles of solute / volume of solution in Litre

therefore

Molarity of OCl^- = 0.00385 mol / 0.0699 L

                         = 0.055 mol/L

                         = 0.055 M

Hence Molarity of OCl^- = 0.055 M

(d) We know

Number of moles = mass of substance in g / gram molar mass of substance

therefore

mass of NaOCl= Number of moles of NaOCl x gram molar mass of NaOCl

mass of NaOCl = 0.00385 mol x 74.5 g /mol

                         = 0.287 g

Hence mass of NaOCl = 0.287 g