Question

1- In the titration between potassium iodate and the sodium thiosulfate solution, if the titration is not performed immediately after the addition of the sulfuric acid, how would this likely affect the calculated concentration of your diluted sodium thiosulfate solution?

2- Why did the titrations performed need to be started immediately after adding the sulfuric acid?

3- A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach solution used instead of pipetting a certain volume of bleach. The student weighed out 0.634 g of commercial bleach solution. It was found that it required 13.24 mL od 0.0732 M sodium thiosulfate solution to react with the iodine produced. What is the percentage of sodium hypochlorite in this bleach sample?

4- The active ingredient in many commercial liquid bleaches is sodium hypochlorite. The bottle lists the percentage of sodium hypochlorite as 6.0%. If the density of commercial bleach is 1.084 g/mL of 0.150 M sodium thiosulfate is required to reach the end point in a titration similar to the one performed in this experiment, if a student analyzed a 2.0 mL sample of bleach.

Answer 1

Answer1) The reaction of IO32- with KI it releases I2 gas ,when H2SO4 is added, so to prevent I2 from evaporating ,it is to be titrated immediately.I2 gas is not very soluble in water, but is soluble only as it forms complex I3- with excess KI.

IO3-+5I- +6H+↔3I2+3H2O

If I2 is evaporated to some extent ,then Na2S2O3 used up for titration will be less, giving an error in the standardization result of Na2S2O3.

2) I2 is not very soluble in water, but is soluble only as it forms complex I3- with excess KI.

IO3-+5I- +6H+↔3I2+3H2O

so to prevent I2 from evaporating ,it is to be titrated immediately

3)ClO-+2I-+H2O→I2+Cl-+2OH-

I-+I2=I3-

I3- (starch) +2S2O32-→3I-+S4O62- +starch

……………………………………………………………………………………….

Overall reaction= ClO-+2S2O32-+H2O→Cl-+S4O62- +2OH-

Molar ratio of hypochlorite: thiosulphate=1:2

So M1V1(hypo)/M2V2(thio)=1:2

Given

M2=0.0732 M sodium thiosulfate

V2=13.24 mL

Moles of thiosulfate=M2V2=0.0732 M*13.24 /1000 L=0.009692 moles

Moles of hypo/moles of thio=1/2

Or,moles of hypo=0.5 *moles of thio=0.5*0.009692 moles=0.00048458 moles

Mass of sodium hypochlorite=moles*molar MASS=0.00048458 moles* 74.5 g/mol=0.0361 g

[NaClO=23+35.5+16=74.5 g/mol]

% sodium hypochlorite=0.0361 g/0.634 g*100=5.7%