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?

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

To find the molarity of HBr and HCl in the original solution, we need to calculate the number of moles of each acid present and divide it by the total volume of the solution.

Let's start by finding the moles of NaOH used in the titration:

Moles of NaOH = Molarity of NaOH × Volume of NaOH used
= 0.1290 mol/L × 0.04756 L
= 0.00614044 moles

Since the balanced chemical equation between NaOH and HCl (or HBr) is 1:1, we can say that the moles of NaOH used are equal to the moles of acid present in the solution.

Now, let's calculate the moles of AgCl and AgBr obtained from the precipitation reaction:

Moles of AgCl = Mass of AgCl / Molar mass of AgCl
= 0.9849 g / 143.32 g/mol (molar mass of AgCl)
= 0.00686788 moles

Moles of AgBr = Mass of AgBr / Molar mass of AgBr
= 0.9849 g / 187.77 g/mol (molar mass of AgBr)
= 0.00523605 moles

Since both AgCl and AgBr are formed from Cl- and Br- ions respectively in a 1:1 ratio, we can say that the moles of AgCl and AgBr are equal to the total moles of Cl- and Br- ions, respectively.

Now, let's find the moles of HBr and HCl:

Moles of HBr = Moles of AgBr
Moles of HCl = Moles of AgCl

Finally, let's calculate the molarity of HBr and HCl:

Molarity of HBr = Moles of HBr / Volume of solution (in L)
= Moles of AgBr / Volume of solution (in L)

Molarity of HCl = Moles of HCl / Volume of solution (in L)
= Moles of AgCl / Volume of solution (in L)

The volume of the solution is given as 100.0 mL, so we need to convert it to liters:

Volume of solution (in L) = 100.0 mL / 1000 mL/L
= 0.1000 L

Substitute the calculated values into the equations to find the molarity of HBr and HCl.