Ask Question
7 September, 08:31

A 10.8ml sample of sulfuric acid titrated with 80.0 ml of 0.200 m mg solution. What is the concentration of the sample given the acid-base reaction shown below

+4
Answers (1)
  1. 7 September, 10:22
    0
    1.48 M

    Explanation:

    Step 1:

    The balanced equation for the reaction. This is given below:

    Mg + H2SO4 - > MgSO4 + H2

    Step 2:

    Determination of the number of mole of Mg in 80.0 mL of 0.200 M Mg solution. This is illustrated below:

    Molarity of Mg = 0.200 M

    Volume of solution = 80 mL = 80/1000 = 0.08L

    Mole of Mg = ?

    Molarity = mole / Volume

    0.2 = mole / 0.08

    Mole = 0.2 x 0.08

    Mole of Mg = 0.016 mole.

    Step 3:

    Determination of the number of mole of H2SO4 that reacted. This is illustrated below:

    Mg + H2SO4 - > MgSO4 + H2

    From the balanced equation above,

    1 mole of Mg reacted with 1 mole of H2SO4.

    Therefore, 0.016 mole of Mg will also react with 0.016 mole of H2SO4.

    Step 4:

    Determination of the concentration of the acid.

    Mole of H2SO4 = 0.016 mole.

    Volume of acid solution = 10.8 mL = 10.8/1000 = 0.0108 L

    Molarity = ?

    Molarity = mole / Volume

    Molarity = 0.016/0.0108

    Molarity of the acid = 1.48 M

    Therefore, the concentration of acid is 1.48 M
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 10.8ml sample of sulfuric acid titrated with 80.0 ml of 0.200 m mg solution. What is the concentration of the sample given the acid-base ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers