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Question

Question #1

If 23.67 mL of a solution of Ba(OH)₂ requires 15.05 mL of a 1.082 M solution of HNO₃ for complete titration, what is the molarity of Ba(OH)₂ solution?  answer in units of M 

 

Question #2

By titration, 26.5 mL of aqueous H₂SO₄ neutralized by 16.6 mL of 0.0385 M LiOH solution. What is the molarity of the aqueous acid solution? answer in units of M 

 

Question #3

A 27.3 mL sample of a solution of RbOH is neutralized by 16.8 mL of a 1.903 M solution of HBr. What is the molarity of the RbOH solution? answer in units of M 

Answer 1

I'll answer question 1 -- follow the same procedure and see if you can answer #2 and #3.

 

First, write the balanced equation:

 

2 HNO₃ + Ba(OH)₂ →  Ba⁺²_aq + NO^3-_aq + 2 H₂O

 

Figure moles of nitric acid used:  15.05 mL HNO₃ x (1 L / 1000 mL) x (1.082 mole/L) =  0.016 moles HNO₃

 

Use the stoichiometric ratio in the eqn:  0.016 moles HNO₃ x ( 1 mole Ba(OH)₂/ 2 moles HNO₃) =

             8.142 x 10^-3 moles Ba(OH)₂

 

Figure the concentration of the Ba(OH)₂:  (8.142 x 10^-3 moles Ba(OH)₂ / 23.67 mL) x (1000 mL/L) =                                         0.344 M  Ba(OH)₂  

 

check:  23.67 mL x (1 L / 1000 mL) x (0.344 mole/L) = 8.143 x 10_-3 moles 

 

So, the answer is that the concentration of the Ba(OH)₂ soln is 0.344 M

 

Use the same approach for numbers 2 and 3.