Ask Question
17 June, 16:24

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?

+1
Answers (1)
  1. 17 June, 16:29
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers