Question

Titration of a 25.00ml KOH (bade) solution required 200.00ml of 0.0050 M acetic acid. What is the molarity of the KOH solution?

Answer 1

Answer:

The molarity of KOH used is 0.04 M

Explanation:

Given base used is KOH and the acid used is Acetic Acid

According to titration;

M1V1 = M2V2

Where, M1 is the molarity of acid

V1 is the volume of acid

M2 is the molarity of base

V2 is the volume of base

Given,  

Volume of KOH (base) – V2 = 25 mL

Volume of acetic acid (acid) used – V1 = 200 mL

Molarity of acetic acid M1 = 0.0050 M

Substituting the values;

0.0050 × 200 = M2 × 25

$\frac{0.0050 \times 200}{25}$

M2 = 0.04 M

The molarity of KOH used is 0.04 M