. A 100.0 mL solution containing HCl and HBr was titrated with 0.1235 M NaOH. Th

Question

. A 100.0 mL solution containing HCl and HBr was titrated with 0.1235 M NaOH. The volume of the base required to neutralize the acid was 47.14 mL. Aqueous silver nitrate was then added to precipitate the Cl- and Br- ions as AgCl and AgBr. The mass of silver halides obtained was 0.9974 g. What are the molarities of HCl and HBr in the original solution?

Explanation / Answer

HBr and HCl are both monoprotic and react in the same ratio (1:1) with NaOH HBr NaOH ---> NaBr + H2O HCl NaOH ---> NaCl + H2O So the moles of NaOH used will be the same as total moles of HBr + HCl in the mixture moles NaOH = molarity x litres = 0.1200 M x 0.04726 L = 0.0056712 mol Therefore moles HCl + moles HBr = 0.00561712 mol There is one Cl- ion in each HCl and one Br- ion in HBr Therefore total moles Br- + Cl- = total moles acid = 0.0056712 mol And, each AgBr and AgCl has only 1 Br- or Cl- so... moles AgCl + moles AgBr = 0.0056712 mol Now, we need to find what combination of moles of AgBr and AgCl = 0.9983 g let moles AgBr = a moles AgCl = total moles - moles AgBr = 0.0056712 - a mass AgBr = molar mass x moles = 187.77 g/mol x a mol = 187.77a mass AgCl = molar mass x moles = 143.32 g/mol x (0.0056712 -a) = 0.812796384 - 143.32a We know the total mass of Halide = 0.9983 g and we know that mass AgBr + mass AgCl = 0.9983 g so 187.77a + 0.812796384 - 143.32a = 0.9983 g solve for a 187.77a - 143.32a = 0.9983 - 0.812796384 44.45 a = 0.185503616 a = 0.185503616 / 44.45 = 0.0041733 mol a = moles AgBr moles AgBr = 0.0041733 mol each AgBr has 1 Br-, so moles Br- (and thus HBr) = 0.0041733 mol moles AgCl = total moles - moles AgBr = 0.0056712 mol - 0.0041733 mol = 0.0014979 mol each AgCl has 1 Cl- so moles Cl- (thus HCl) = 0.0014979 mol total volume was 100.0 ml = 0.1000 L molarity = moles / Litres molarity HBr = 0.0041733 mol / 0.1000 L = 0.04173 M molarity HCl = 0.0014979 mol / 0.1000 L = 0.01498 M

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