what is the molarity of a solution of sodium hydroxide if it 15.0 mL requires 25.0 ml of 0.25M HCl to reach the end point

0.42 M

Explanation:

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

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of acid, HCl (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 1

Finally, we shall determine the molarity of the base, NaOH. This can be obtained as follow:

Volume of acid, HCl (Vₐ) = 25 mL

Molarity of acid, HCl (Mₐ) = 0.25 M

Volume of base, NaOH (V₆) = 15 mL

Molarity of base, NaOH (M₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.25 × 25 / M₆ × 15 = 1

6.25 / M₆ × 15 = 1

Cross multiply

6.25 = M₆ × 15

Divide both side by 15

M₆ = 6.25 / 15

M₆ = 0.42 M

Therefore, the molarity of the base, NaOH is 0.42 M

it should be p1 v1 = p2 v2

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