Question

Bleaching powder and bleach solutions are produced on a large scale and are used in several household products. The effectiveness of bleach solution is often measured by iodometry. 25 ml of household bleach solution was mixed with 30 ml of 0.50M KI and 10 ml of 4N acetic acid. In the titration of the liberated iodine, 48 ml of 0.25N $N{a_2}{S_2}{O_3}$ was used to reach the end point. The molarity of the household bleach solution is
  (A) 0.48M
  (B) 0.96M
  (C) 0.24M
  (D) 0.024M

Answer 1

Hint: First calculate the equivalents of ${S_2}{O_3}^{ - 2}$ involved and then use the chemical law of equivalence to find the equivalents of bleach used. This will lead us to its concentration in terms of normality, so convert it to molarity terms.

Complete answer:
-The reactions involved here are:
                $2C{H_3}COOH + CaOC{l_2} \to {(C{H_3}COO)_2}Ca + {H_2}O + C{l_2}$
                                             $C{l_2} + 2KI \to 2KCl + {I_2}$
                    $2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$

-Here we will be using the law of chemical equivalence.
According to this law whenever two substances react, the equivalents of one will be equal to the equivalents of the other and the equivalents of any product will be equal to the equivalents of the reactant.
For the reactions written above this law can be written as:
     eq. of ${I_2}$ = eq. of $C{l_2}$ = eq. of ${S_2}{O_3}^{ - 2}$ = eq. of bleach ($CaOC{l_2}$) = eq. of $C{H_3}COOH$ (1)

-The question has given us the N and V of ${S_2}{O_3}^{ - 2}$ to be 0.25 N and 48 ml.
So, first we will calculate the equivalents of ${S_2}{O_3}^{ - 2}$:
                                            eq. = N × V (ml) (2)
                                                  = 0.25 × 48
                                                  = 12 m.eq.

-Now from (2) we can say that: eq. of ${S_2}{O_3}^{ - 2}$ = eq. of $CaOC{l_2}$
So, the eq. of $CaOC{l_2}$ = 12 m.eq.
We also know that: (eq. = N × V) and (N = M × n-factor)
So, from this we can say that: eq. = M × n-factor × V (3)
For $CaOC{l_2}$ n-factor = 2

Hence we will now calculate the molar concentration of $CaOC{l_2}$ from equation (3):
                                                eq. = M × n-factor × V
                                                 12 = M × 2 × 25
                                               $M = \dfrac{{12}}{{2 \times 25}}$
                                                       = 0.24 M
So, the correct answer is “Option C”.

Note: The law of chemical equivalence if the fundamental basis for all acid-base titrations. So, according to it at the end point of titration, the volume of 2 titrants reacted with the same number of equivalents or milli equivalents.
Also, the household bleach has its properties due to the presence of chlorine which is a powerful oxidizer. They have broad spectrum bactericidal properties, which makes it highly useful for disinfecting, sterilizing and sanitizing purposes.