Answer to Question #101232 in Physical Chemistry for SANI YAHAYA

Question

The quantity of chloride ion in water supply was determined by the titrating the sample against AgNO3 solution.
AgNO3(aq) + Cl- = AgCl + NO3

i. what & of Cl- is present in 10g sample of water, if 20.2cm3 of 0.1M of AgNO3 is required to react completely with all the Cl- IN THE SAMPLE?
ii. express the quantity of Cl- present in ppm

Answer 1

$AgNO_3(aq) + Cl^-\to AgCl + NO_3$

(i) The stochiometric coefficient of silver nitrate and chloride is in ratio 1:1

so number of chloride ions present in the 10 gram of water sample = number of particles of silver nitrate required

moles of chloride ions present = 20.2*10^-3*0.1 =2.02*10^-3

(ii) moles of water in grams of water = 10/18

number of particles of water in 10 grams of water = (10/18)*Na (Na is the Avogadro number )

number of particles of chloride ions in 10 gram of water sample = 2.02*10^-3*Na

ratio of particles of chloride ions to water molecules

= (2.02*10^-3*Na)/((10/18)*Na)

= 36.36*10^-4

converting this ratio into ppm by multiplying by 1 million(1000000)

we get 36.36*10³