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Question

Question #1

In a titration 23.46 ml of 0.00325 M Ba(OH)₂ neutralized 13.8 ml of HCL solution. What is the molarity of the HCl solution?

Answer in units of M

 

 

Question #2

 Find the molarity of a Ca(OH)₂ solution, given that 287 mL of it is neutralized in a titration by 176 mL of 0.0066M HNO_3. 

Answer in units of M 

 

 

 Question #3

In a titration experiment, a 28.2 mL sample of 0.0106M Ba(OH)₂ just neutralized 14.5 mL of HNO₃ solution. Calculate the molarity of the HNO₃ solution.

Answer in units of M

Answer 1

I will do Question 1.  Follow the same procedure and you should be able to do Question 2 and Question 3.

 

First, write the balanced equation for the titration reaction:

 

Ba(OH)₂ + 2HCl →  2 BaCl + 2 H₂O

 

At the equivalence point for this reaction, the number of moles of OH^- delivered neutralized an equivalent number of moles of H⁺

 

and the relationship M₁V₁ = M₂V₂ is followed.

 

Using the stoichiometry of the reaction, solve for M2:

 

23.46 mL Ba(OH)₂ x (1L/1000 mL) x .00325 mole Ba(OH)₂/L x (2 mole OH^-/mole Ba(OH)₂) x (2 mole HCl/ 2 mole OH^-) x (1/13.8 mL HCl) x 1000 mL HCl/L HCl) = 0.011 M HCl

 

∴ the concentration of the HCl is 0.011 M