In standardization of NaOH solution, a student found that 25.55cm3 of base neutralized exactly 21.35cm3 of 0.12M HCl. Find the molarity of the NaOH.

0.1M

Explanation:

We'll begin by writing a balanced equation for the reaction between HCl and NaOH. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the above equation, we obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 1

Data obtained from the question include:

Ma (Molarity of the acid ) = 0.12M

Va (volume of the acid) = 21.35cm3

Mb (Molarity of the base) =?

Vb (volume of the base) = 25.55cm3

Using the equation MaVa/MbVb = nA/nB, the molarity of the base can be obtained as follow:

MaVa/MbVb = nA/nB

0.12 x 21.35 / Mb x 25.55 = 1

Cross multiply to express in linear form.

Mb x 25.55 = 0.12 x 21.35

Divide both side by 25.55

Mb = (0.12 x 21.35)/25.55

Mb = 0.1M

Therefore, the molarity of the base (NaOH) is 0.1M

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