Toxic cyanide ions can be removed from wastewater by adding hypochlorite: 2CN^-
ID: 921309 • Letter: T
Question
Toxic cyanide ions can be removed from wastewater by adding hypochlorite: 2CN^- (aq) + 50Cl^- (aq) + H_2O (l) rightarrow N_2(g) + 2HCO3^- (aq) + 5Cl^- (aq) If 1.50 Time 10^3 L of 0.125 M OCl^- are required to remove the CN^- in 3.4 Times 10^6 L of wastewater, what is the CN^- concentration in the water in mg/L? How many mL of 0.575 M AgNO3 would you need to add to a 50.00 mL aliquot of the final solution (consider the volumes simply additive) to precipitate the chloride ions formed in the reaction?Explanation / Answer
Solution :-
a)1.50*10^3 L of 0.125 M OCl^-
3.4*10^6 L waste water
Concentration of the CN- =?
Lets first calculate the moles of the OCl-
Moles = molarity * volume in liter
= 0.125 mol per L * 1.50*10^3 L
= 187.5 mol OCl-
Now lets calculate the moles of the CN- using the mole ratio of the OCl- and CN-
187.5 mol OCl- * 2 mol CN- / 5 mol OCl- = 75 mol CN-
Now lets calculate the concentration of the CN-
[CN-] = moles / volume
= 75 mol / 3.4*10^6 L
= 2.08*10^-5 M
So the concentration of the CN- = 2.08*10^-5 M
b) moles of Cl- = 187.5 mol since mole ratio of the OCl- and Cl- is 1 : 1
so the molarity of the Cl- at total volume is
total volume = 1.50*10^3 L + 3.4*10^6 L = 3401500 L
so concentration of the Cl- = moles / volume
= 187.5 mol / 3401500 L
= 5.512*10^-5 M
Now lets calculate the volume of the AgNO3 needed
Volume of AgNO3 = molarity of Cl- * volume of Cl- / molarity of AgNO3
= 5.512*10^-5 M * 50.00 ml / 0.575 M AgNO3
= 0.00479 L
converting liter to ml
0.00479 L * 1000 ml / 1 L = 4.97 ml AgNO3 so we can round it to 5.00 ml AgNO3
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.