Can someone please help me answer these 2 questions for me I would really apprec

Question

Can someone please help me answer these 2 questions for me I would really appreciate I'am very lost.

Thank you

Introduction In this experiment you are going to determine the oxidizing capacity of commercial bleach. The hypochlorite ion (CIO) is one of the most commonly used active ingredients in commercial bleach for making clothes whiter and removing stains. The CIO ion is an oxidizing agent which removes electrons from the "stain" molecules and henc visible light e destroying substances that absorb Equation 1 When light is no longer absorbed, the material appears whiter. Experimentally, we will use two oxidation-reduction reactions to determine the oxidizing capacity of a bleach sample. First, an excess of potassium iodide (KI) is added to a solution containing the bleach sample. The CIO ion in the bleach will oxidize the iodide ion to iodine according to the following reaction: CIO' + 2H+ + 2e. 2t CT+H20 12 + 2 e Equation 2 H will be present in excess. The acid in this experiment will be H2SO4. The iodine (a) produced in the solution will then be determined by titration with a standardized thiosulfate solution which reduces iodine stoichiometrically according to the following reaction: 252032- 12 +2 e. S4062. +2e' 2r Equation 3 Equation (2) tells us that, for every mole of CIo reacted one mole of 12 is produced. Equation (3) indicates that it takes two moles of S20,2 to react with one mole of L. Adding Eq. (2) and (3) will give an overall stoichiometric relationship between ClO in the bleach sample and S,O,'- Equation 4 At the equivalence point of the titration: Equation 5 moles of cio1 moles of s203 2

Explanation / Answer

Answer:

Reaction of chlorine gas with Sodium hydroxide will produce sodium hypochloride as follows,

Cl2 +2NaOH --> NaOCl + NaCl + H2O-------------------------------------------------------1

ClO- + H2O + 2e- --> Cl- + 2OH- -------------------------------------------------------------2

in an experiment to determine oxidizing capasity of bleach it is first reacted with excess of KI. In this reaction ClO- ion oxidises to Cl- and iodine ion reduce to iodine gas as follows,

ClO- + H2O + 2e- --> Cl- + 2OH-

+                         2I- --> I2 + 2e-

OCl-(aq) + 2H+(aq)   + 2I-(aq)     I2(aq)   + Cl- (aq)+ H2O(l)---------------------------3

From equation 1 we can say that we need one molecule of chlrine gas and 2 molecules of NaOH to produce one molecule of bleach.

So number of moles of bleach in 2 kg of bleach = 2000 g/74.44 = 26.86 moles

26.86 moles of NaoCl dissociates completely in the water to give chloride ion and OH- ion

So at equilibrium concnetration of chloride ion will be 26.86 mols/L of water

Moles of chlorine gas = 2*26.86 mols/L = 53.73

Now we can calculate volume of gas at stp as follows,

PV=nRT

V = nRT/P = (53.73 × 8.314×10-2 Lbar K1mol1×298K)/1 bar

V= 1331 Liter

Therefore, 1331 Liter of chlorine gas is require to produce 2 Kg of Sodium hypochloride

Answer 5:

Given that bleach is 6% by weight of sodium hypochloride

Density of bleach is = 1.07 g/ml

Bleach is 6% of the Weight of sodium hypochloride = 6/100*

Weight of the soium hypochloride in the bottle = 6 g/100 g*1.07 g*L-1 = 0.0642 gm of bleach per ml

So weight of bleach per liter = 64.2 g

Molarity of 1 liter of bleach = 64.2gm /77.44 = 0.8290M

This will dissociate completely to give HOCl- and OH-

Therefore concnetration of HOCl will be 0.8290M

NaOCl + H2O --> HOCl + OH-

Now HOCl + H2O --> H3O+ + OCl-

Hypochlorous acis is a weak acid forming the hypocholite ion, OCl-, when it dissociates in water according to the equation above.

Now, Ka = 3.5 x10 -8

Ka= [H3O+][OCl-]/[HOCl]

3.5 x10 -8 = x2/0.8290

X= 1.703 ×10-4

[H3O+]=[H+]= 1.703 ×10-4M

pH = -log 1.703 ×10-4M

pH= 3.7

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