Question

A 100.0mL solution containing aqueous HCl and HBr was titrated with 0.1290M NaOH. The volume of base required to neutralize the acid was 47.56mL. Aqueous AgNO3 was then added to precipitate the Cl- and Br- ions as AgCl and AgBr. The mass of the silver halides obtained was 0.9849 g.

a.)What is the molarity of HBr in the original solution?

b.)What is the molarity of HCl in the original solution?

Answer 1

Two equations in two unknowns.
Let X = mols HBr
and Y = mols HCl
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equation 1 and 2 follow:
X + Y = 0.1290 x 0.0475
X(molar mass AgBr) + Y(molar mass HCl) = 0.9849
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Solve for X and Y, then M HCl = mols HCl/L and M HBr = mol HBr/L