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telo118 [61]
3 years ago
14

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
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
3 0

Answer:

0.200L of 0.300 Al(OH)₃ are necessaries

Explanation:

A diprotic acid, H₂X, reacts with aluminium (III) hydroxide, Al(OH)₃ as follows:

3 H₂X + 2 Al(OH)₃ →  6H₂O + Al₂X₃

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:

<em>Moles H₂X:</em>

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

<em>Moles Al(OH)₃:</em>

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

<em>Volume 0.300M Al(OH)₃:</em>

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

<h3>0.200L of 0.300 Al(OH)₃ are necessaries</h3>
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Explanation:

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