Question

a 350 mL vol of .15 M Ba(OH)2 was completely neutralized by 35.0 mL of HCl. What is the molarity of the HCl solution?

Answer 1

For complete neutralisation -

Moles of OH^- = Moles of H⁺

Ba(OH)₂  ------ > Ba²⁺  + 2OH^-

From the chemical equation above it is clear that 1 mole of Ba(OH)₂ gives 2 mole of OH^- .

Molarity*Volume = Moles

Molarity of Ba(OH)₂ = 0.15M (Given)

Volume of Ba(OH)₂ used for neutralisation = 350mL

Millimoles of Ba(OH)₂ = 0.15*350   

Millimoles of OH^- obtained from Ba(OH)₂ = 2*(0.15*350)

HCl -----> H⁺  + Cl^-

According to the equation above it is clear that 1 mole of HCl gives 1 mole of H⁺.

Volume of HCl used for neutralisation = 35.0mL (Given)

Let us take molarity of HCl be x.

Molarity*Volume = Moles

Millimoles of HCl = x*35.0

Millimoles of H⁺  = x*35.0

For complete neutralisation millimoles or moles of OH^-  should be equal to millimoles or moles of H⁺.

2*(0.15*350) = x*35.0

x = 3M

Molarity of HCl is 3M.

Note-Since we have not changed volume to litres so we are getting millimoles instead of moles.