Question

Concentrated hydrochloric acid solution is 37.0% HCL and has a density of 1.19g/ml. A dilute solution of HCL is prepared by diluting 4.50mL of this acid to 100.00mL with water. Then 10.0mL of the dilute HCL is used for the reaction with AgNO3 solution as shown below: HCL(aq)+AgNO3(aq) to HNO3(aq)+AgCL(s) How many mL of 0.1105M AgNO3 solution is required to precipitate all of the chloride as AgCl(s)? Please show how you get the answer. Thank you!

Answer 1

37% of HCl means 37g of HCl dissolve in 100ml of solution = 37/100 = 0.37g/ml

density of HCl = 1.19gm/ml

valume of HCl= 4.5ml

mass = volume*density

mass of 37% of HCl       = 4.5*1.19*0.37 = 1.98g

molar mass of HCl = 36.5g/mole

no of moles of HCl = W/GMM

                               = 1.98/36.5 = 0.0542 moles

molarity of HCl in 4.5ml = no of moles of HCl/volume in liters

                                        = 0.0542/0.0045 = 12M

the 12M HCl was diluted to 100ml

before dilution molarity(M1) = 12M

                      volume (V1)    = 4.5ml

after dilution molarity( M2 ) =

                    volume             = 100ml

                     M1V1         = M2V2

                      M2 = M1V1/V2 = 12*4.5/100 = 0.542M

the no of moles of HCl in 10ml of sample = 0.542*0.01 = 0.00542 moles

1 mole of HCL react with 1 mole of AgNO3

HCl +AgNO3\rightarrowAgCl +HNO3

no of moles of HCl = no of moles of AgNO3

so no of moles of AgNO3 = 0.00542 moles

molarity = no of moles/volume in liters

volume in liters = no of moles/molarity

                           = 0.00542/0.1105 = 0.0491liters = 49.1ml

49.1 ml of AgNO3 is required to pricipitate of AgCl