Question

A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammonia. What is the concentration of the original ammonia solution?

Answer 1

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M