Question

Chemistry 50 Worksheet Unit 4: Solubility Equilibrium 9. What mass of Cl is required to precipitate all the silver ions from 50.0 mL of a 0.0500 M solution of AgNO3? Note: the system is NOT at equilibrium. (3 marks)

Answer 1

- AgNO₃ + ct - Agel L + NO3 AgNO3 exists in solution as Agt and Nog. el ions reacts with Agt and forms Agce Precipitation. from the reaction we can see for £M of Imale AgNO3 we need a mole of et Let's find the number of moles of AgNO3 Presents in 50.0 mL of a 0.0500 M solution. No of moles Molarity = volume in liters i No of moles - Molarly x rolume (in liters) = 0.0500 may x 500 1ooo 2.5 - 7000 mol = 0.0025 mol So, we found that 0.0025 mols of AgNO3 Presents in the solution. To completely precipitate Agt, we need minimum 0.0025 mols of an Weight Noof molef = Molecular weight in Weight = 0.0025 molx 35.5 g/mol [Molecular weight = 0.08875g of chlorine = 35.5g/m] = 88.75ing & we need 88.75 mg of ce- (answer)