Question

$ 5.0g $ of bleaching powder was suspended in water and volume made up to half a liter. $ 20ml $ of this suspension when acidified with acetic acid and treated with excess potassium iodide solution liberated iodine ,which required $ 20ml $ of a Deci-normal hypo solution for titration. Calculate the value of $ 10 \times x $ , where x is the percentage of available chlorine in bleaching powder?

Answer 1

Hint: Write down the chemical reaction and calculate the equivalent mass of chlorine and then relate them using $ {N_1}{V_1} = {N_2}{V_2} $ after find out the mass of chlorine and then find the percentage.

Complete answer:
Let us first write down the chemical reaction for this question. In question it is given that bleaching powder is suspended in water to make up to half a liter of volume. For this step the reaction will be:
 $ CaOC{l_2} + {H_2}O \to Ca{(OH)_2} + C{l_2} $
Now this suspension was acidified and treated with excess of potassium iodide then the reaction will become:
 $ C{l_2} + 2KI \to 2KCl + {I_2} $
Now this product is titrated against a hypo solution then this will become:
 $ {I_2} + 2N{a_2}{S_2}{O_3} \to N{a_2}{S_4}{O_6} + NaI $
Now using the concept of equivalence
 $ {\text{1 mole CaOC}}{{\text{l}}_2} = 1{\text{ mole C}}{{\text{l}}_2} = 1{\text{ mole }}{{\text{I}}_2} = 2{\text{ mole N}}{{\text{a}}_2}{S_2}{O_3} $
Now we will find the equivalent mass of chlorine
 $ C{l_2} = \dfrac{{71}}{2} = 35.5 $
And $ {N_1}{V_1}(C{l_2}{\text{ in bleaching powder) = }}{{\text{N}}_2}{V_2}(hypo solution) $ , now solving it we will get:
 $ {N_1} \times 20 = 0.1 \times 20 $
 $ {N_1} = 0.1N $
Now the normality of $ CaOC{l_2} $ solution will be $ 0.1N $
Now we will find out the amount of chlorine in the pure sample.
It will be $ 0.1 \times 35.5g{L^{ - 1}} = 3.55g{L^{ - 1}} $
The amount of impure sample in chlorine will be H
 $ \dfrac{{5.0g}}{{0.5L}} = 10g{L^{ - 1}} $
So the pure percentage of pure chlorine (x) available in the bleaching powder is :
 $ \dfrac{{3.55}}{{10}} \times 100 = 35.5\% $ this is our value of x
But in question it is asked to find the value of $ 10 \times x $ so its value will be $ 10 \times 35.5 = 355 $ which our answer is.

Note:
While using the concept of equivalence keep in mind that the equations you are using should be balanced for easiness and all the units should be in a system like if you are taking volume in liter then everywhere it should be in liter.