Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.



What volume of 0.062 M HCl is needed to react to completely neutralize 19.4 mL of 0.050 M Ba(OH)2? *


Sagot :

Answer:

31.3 mL

Explanation:

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

2HCl + Ba(OH)₂ —> BaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nₐ) = 2

The mole ratio of the base, Ba(OH)₂ (n₆) = 1

Finally, we shall determine the volume of the acid need for the reaction as follow:

Molarity of acid, HCl (Mₐ) = 0.062 M

Mole ratio of the acid, HCl (nₐ) = 2

Volume of base, Ba(OH)₂ (V₆) = 19.4 mL

Molarity of base, Ba(OH)₂ (M₆) = 0.050 M

Mole ratio of the base, Ba(OH)₂ (n₆) = 1

Volume of acid, HCl (Vₐ) =?

MₐVₐ / M₆V₆ = nₐ / n₆

0.062 × Vₐ / 0.050 × 19.4 = 2/1

0.062 × Vₐ / 0.97 = 2

Cross multiply

0.062 × Vₐ = 0.97 × 2

0.062 × Vₐ = 1.94

Divide both side by 0.062

Vₐ = 1.94 / 0.062

Vₐ = 31.3 mL

Thus, the volume of the acid needed is 31.3 mL