Question

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).

Answer 1

Answer:

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

Answer 2

Answer:

0.2M

Explanation:

KOH(aq) + HCl(aq) ⇒  KCl(aq) + H2O(l)

We express the moles of analyte (HCl) and titrant based (KOH) on their molar concentration:

M1 * V1 = M2 * V2

The molarity of the solution is calculated with the following equation:

M2 = V1 x M1 / ​​V2

Where:

V2 = valued sample volume

V1 = volume of titrant consumed (measured with the burette)

M1 = concentration of titrant solution

M2 = concentration of sample

M2 = 100mL * 0.026M / 13mL = 0.2M