What is the molarity of a sodium hydroxide solution if 24 mL of the solution
is titrated to the end point with 12 mL of 0.50M sulfuric acid? (Remember:
MaVa = MbVb)
•0.10M
•0.25M
•0.50M
•0.75M

Option 3. 0.50M

Explanation:

Step 1 :

We'll begin by writing the balanced equation for the reaction. This is given below:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

From the above equation, we obtained the following:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 2

Step 2:

Data obtained from the question. This includes the following:

Molarity of base (Mb) =?

Volume of base (Vb) = 24 mL

Volume of acid (Va) = 12 mL

Molarity of acid (Ma) = 0.50M

Step 3:

Determination of the molarity of the base.

The molarity of the base can be obtained by applying the following equation:

MaVa/MbVb = nA/nB

0.5 x 12/Mb x 24 = 1/2

Cross multiply to express in linear form

Mb x 24 = 0.5 x 12 x 2

Divide both side by 24

Mb = (0.5 x 12 x 2) /24

Mb = 0.5 M

Therefore, the molarity of NaOH is 0.5 M

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