In a titration of HCl with NaOH, 100mL of the base was required to neutralize 20mL of 5.0 M HCl. What is the molarity of the NaOH?

1 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Volume of base (Vb) = 100mL

Volume of ac (Va) = 20mL

Molarity of acid (Ma) = 5M

Molarity of base (Mb) = ... ?

Step 2:

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

HCl + NaOH - > NaCl + H2O

From the above equation, the following were obtained:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 3:

Determination of the molarity of the base.

The molarity of the base can be obtained as follow:

MaVa/MbVb = nA/nB

5 x 20 / Mb x 100 = 1

Cross multiply to express in linear form

Mb x 100 = 5 x 20

Divide both side by 100

Mb = (5 x 20) / 100

Mb = 1 M

Therefore, the molarity of the base is 1 M