Ask Question
3 June, 21:15

In a titration how much 0.50 M HCl is needed to neutralize 1 liter of a 0.75 M solution of NaOH?

1.5 L

2.25 L

1.0 L

0.75 L

+1
Answers (1)
  1. 3 June, 22:42
    0
    1.5 L

    Explanation:

    Step 1:

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

    HCl + NaOH - > NaCl + H2O

    From the balanced equation above, The mole ratio of the acid (nA) = 1

    The mole ratio of the base (nB) = 1

    Step 2:

    Data obtained from the question. This includes the following:

    Molarity of acid (Ma) = 0.5M

    Volume of acid (Va) = ... ?

    Volume of base (Vb) = 1L

    Molarity of base (Mb) = 0.75M

    Step 3:

    Determination of the volume of the acid needed for the reaction.

    Using the formula:

    MaVa/MbVb = nA/nB

    The volume of the acid needed for the reaction can be obtained as follow:

    0.5 x Va / 0.75 x 1 = 1

    Cross multiply

    0.5 x Va = 0.75 x 1

    Divide both side by 0.5

    Va = 0.75 / 0.5

    Va = 1.5 L

    Therefore, the volume of the acid, HCl needed for the reaction is 1.5L
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “In a titration how much 0.50 M HCl is needed to neutralize 1 liter of a 0.75 M solution of NaOH? 1.5 L 2.25 L 1.0 L 0.75 L ...” 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