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Irina-Kira [14]
3 years ago
10

Why does the titration of a weak acid with a strong base always have a basic equivalence point? Why does the titration of a weak

acid with a strong base always have a basic equivalence point? At the equivalence point, the base has all been converted into its conjugate acid, resulting in a weakly basic solution. At the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly acidic solution. At the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly basic solution. At the equivalence point, the base has all been converted into its conjugate base, resulting in a weakly basic solution.
Chemistry
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer: Option (c) is the correct answer.

Explanation:

When a weak acid reacts with a strong base then it results into the formation of a basic solution. Hence, the resulting solution will always have a pH greater than 7.

Since, at the equivalence point number of hydrogen ions become equal to the hydroxide ions. Therefore, pH of solution will be about 7.

So at the equivalence point, the weak acid will get neutralized due to the addition of strong base. Therefore, it will lead to the formation of conjugate base.

As a result, the solution will become slightly basic in nature.

Thus, we can conclude that at the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly acidic solution because at the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly basic solution.

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3 years ago
CsH«N2(l) + CuCl(s)
bulgar [2K]

0.498 moles of copper(II) phthalocyanine would be produced by

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<em>Copper(ll) phthalocyanine (Cu(C₃₂H₁₆N₈)) is produced by the cyclotetramerization of phthalonitrile (C₈H₄N₂) according to the following reaction: 4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g) How many moles of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 255 grams of phthalonitrile in the presence of excess copper(II) chloride?</em>

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar mass of C₈H₄N₂ is 128.13 g/mol. The moles corresponding to 255 g of C₈H₄N₂ are:

255 g \times \frac{1mol}{128.13 g} = 1.99 mol

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 1.99 moles of C₈H₄N₂ are:

1.99 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.498 mol Cu(C_{32}H_{16}N_8)

0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

You can learn more about stoichiometry here: brainly.com/question/22288091

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