How many liters of 0.300 M aluminum (III) hydroxide do you need to titrate 0.300 L of a 0.300 M diprotic acid to the equivalence point

To solve this question we must find the moles of the H₂X, using the balanced reaction we can find the moles of Al(OH)₃ and its volumen knowing its molar concentration is 0.300M:

Moles H₂X:

0.300L * (0.300mol / L) = 0.0900moles H₂X

Moles Al(OH)₃:

0.0900moles H₂X * (2mol Al(OH)₃ / 3mol H₂X) = 0.0600 moles Al(OH)₃

Volume 0.300M Al(OH)₃:

0.0600 moles Al(OH)₃ * (1L / 0.300moles) =

0.200L of 0.300 Al(OH)₃ are necessaries

Eduard22sly Eduard22sly · Answer 2 · 2021-12-10T12:02:01+01:00

The volume of Al(OH)₃ needed for the reaction is 0.2 L

We'll begin by writing the balanced equation for the reaction.

3H₂X + 2Al(OH)₃ → 6H₂O + Al₂X₃

The mole ratio of the acid, H₂X (nA) = 3

The mole ratio of base, Al(OH)₃ (nB) = 2

From the question given above, the following data were

Volume of acid, H₂X (Va) = 0.3 L

Molarity of acid, H₂X (Ma) = 0.3 M

Molarity of base, Al(OH)₃ (Mb) = 0.3 M

Volume of base, Al(OH)₃ (Vb) =?

MaVa / MbVb = nA / nB

(0.3 × 0.3) / (0.3 × Vb) = 3/2

0.09 / (0.3 × Vb) = 3/2

Cross multiply

0.09 × 2 = 0.3 × 3 × Vb

0.18 = 0.9 × Vb

Divide both side by 0.9

Vb = 0.18 / 0.9

Vb = 0.2 L

Therefore, the volume of Al(OH)₃ needed for the reaction is 0.2 L

Learn more: https://brainly.com/question/25768579