1.- When 38 mL of 0.1250 M H 2 SO 4 isadded to 125 mL of a solution of BaCl 2 ,

Question

1.- When 38 mL of 0.1250 M H2SO4 isadded to 125 mL of a solution of BaCl2, a precipitateof BaSO4 forms. The BaSO4 is thenfiltered from the solution, dried, and weighed. If the recoveredBaSO4 is found to have a mass of 0.105 g, what was theconcentration of barium ions in the original solution of bariumchloride? Hint: write out a reaction. a) 3.59x10^-4M   b) 1.79x10^-4 M   c) 7.18x10^-3M d) 3.60x10^-3 M   e) 1.79x10^-3 M 2.- Which of these choices is the correct net ionic eqn. forthe reaction that occurs when solutions of Pb(NO3)2 and NH4Br aremixed? a) Pb(NO3)2(aq) + 2NH4Br(aq)-------> NH4NO3(aq) +PbBr2(s) b) Pb2+(aq) + 2Br-(aq) --------->PbBr2(s) c) Pb2+(aq) + 2NO3-(aq)+ 2NH4+(aq) + 2Br-(aq)--------> 2NH4+(aq) +2NO3-(aq) + PbBr2(s) d) NH4+(aq) +NO3-(aq) ---------->2NH4NO3(s) e) No reaction occurs when the solutions are mixed. 1.- When 38 mL of 0.1250 M H2SO4 isadded to 125 mL of a solution of BaCl2, a precipitateof BaSO4 forms. The BaSO4 is thenfiltered from the solution, dried, and weighed. If the recoveredBaSO4 is found to have a mass of 0.105 g, what was theconcentration of barium ions in the original solution of bariumchloride? Hint: write out a reaction. a) 3.59x10^-4M   b) 1.79x10^-4 M   c) 7.18x10^-3M d) 3.60x10^-3 M   e) 1.79x10^-3 M 2.- Which of these choices is the correct net ionic eqn. forthe reaction that occurs when solutions of Pb(NO3)2 and NH4Br aremixed? a) Pb(NO3)2(aq) + 2NH4Br(aq)-------> NH4NO3(aq) +PbBr2(s) b) Pb2+(aq) + 2Br-(aq) --------->PbBr2(s) c) Pb2+(aq) + 2NO3-(aq)+ 2NH4+(aq) + 2Br-(aq)--------> 2NH4+(aq) +2NO3-(aq) + PbBr2(s) d) NH4+(aq) +NO3-(aq) ---------->2NH4NO3(s) e) No reaction occurs when the solutions are mixed.

Explanation / Answer

Moles of BaSO4 = mass/ molar mass                            =0.105 g / 233.4 g/mol                            =4.5 * 10-4 moles . BaCl2 + H2SO4 -------> BaSO4+ 2 HCl . So moles of Barium ions originally present = 4.5 *10-4 moles we know volume of BaCl2 = 125 mL or 0.125 L . Molarity of barium ions = moles / volume                                     =4.5 * 10-4 moles / 0.125 L                                     =3.6 * 10-3 M . When lead nitrate and ammonium bromide are mixed, lead bromidebeing sparingly soluble precipitates out. Hence the net ionic equation is: b) Pb2+(aq) +2Br-(aq) ---------> PbBr2(s)

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