In a titration, 100 mL of 0.026 M HCl (aq) is neutralized by 13 mL of KOH(aq). Calculate the molarity of KOH (aq).

0.2M

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 100mL

Molarity of the acid (Ma) = 0.026 M

Volume of base (Vb) = 13mL

Molarity of the base (Mb) =..?

Step 2:

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

HCl + KOH —> KCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the molarity of the base, KOH. This can be obtained as follow:

MaVa/MbVb = nA/nB

0.026 x100 / Mb x 13 = 1

Cross multiply to express in linear form

Mb x 13 = 0.026 x 100

Divide both side by 13

Mb = 0.026 x 100 / 13

Mb = 0.2M

Therefore, the molarity of the base, KOH is 0.2M

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