Sign In

JEE Advance - Chemistry (2012 - Paper 2 Offline - No. 14)

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
Bleaching powder and bleach solution are produced on a large scale and 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 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is
0.48 M
0.96 M
0.24 M
0.024 M

Explanation

Consider the titration reaction

CaOCl2 + 2KI $$\to$$ I2 + Ca(OH)2 + KCl

According to this reaction, 25 mL of CaOCl2 reacts with 30 mL of 0.50 M KI

I2 + 2Na2S2O3 $$\to$$ Na2S4O6 + 2NaI

Given that 48 mL of 0.25 N Na2S2O3 was used to reach the end point. So, the number of moles of I2 produced = 48 $$\times$$ 0.25/2 = 6.

According to the reaction,

Number of millimoles of bleaching powder = Number of moles of I2 = (1/2) $$\times$$ Number of moles of Na2S2O3 = 6

So, the molarity of CaOCl2 is

$${{Number\,of\,moles\,of\,bleaching\,powder} \over {Volume\,of\,solution}} = {{6\,mmol} \over {25\,mL}} = 0.24M$$

Comments (0)

Advertisement