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

1.5 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