149k views
5 votes
A chemist titrates 0.200 M NaOH, strong base, with 50.00 ML of 0.100 M HCI, strong acid. How many mL of NaOH will be required to titrate to the endpoint

1 Answer

2 votes

Answer:

25 mL of NaOH

Step-by-step 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,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 1

Finally, we shall determine the volume of NaOH required for the reaction. This can be obtained as follow:

Molarity of the base, NaOH (M₆) = 0.2 M

Molarity of the acid, HCl (Mₐ) = 0.1 M

Volume of the acid, HCl (Vₐ) = 50 mL

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 1

Volume of the base, NaOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ / n₆

0.1 × 50 / 0.2 × V₆ = 1

5 / 0.2 × V₆ = 1

Cross multiply

5 = 0.2 × V₆

Divide both side by 0.2

V₆ = 5/0.2

V₆ = 25 mL

Thus, 25 mL of NaOH is needed for the reaction

User Activebiz
by
5.5k points