Question

To a 250.0 mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M AgNO3, 0.245 M NaBr, and 0.100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigorously What mass of AgBr would precipitate from this mixture? (Hint: The Ksp of AgBr is 5.4 x 10-13 and the Ky of Ag(CN)2-is 1.0 × 1021) Number g AgBr

Answer 1

concentration of Ag+ in solution = 0.01 M*1ml/250 ml = 0.00004 M

concentration of Br- in solution = 0.245 M * 1 ml/250 ml = 0.00098 M

moles of Ag+ = 0.00004 M*250 ml = 0.01 mmols

moles of Br- = 0.00098M*250 ml = 0.245mmols

1 mole of AgBr with combine with 1 mole of NaBr to form 1 mole of AgBr

moles of AgBr formed = 0.01 mmols

mass of AgBr formed = 0.01 mmol * 187.77 g/mol/1000 = 1.88 mg