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yuradex
1 month ago
5

What is the conjugate acid of each of the following? What is the conjugate base of each?

Chemistry
1 answer:
lions [2.9K]1 month ago
3 0

Answer:

a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.

b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.

d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.

(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.

g. HS⁻ is amphoteric.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

This is similar to the case of bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hydrazinium acts as an acid, making hydrazine its conjugate base.

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Answer:

The original halide's formula is SrCl₂.

Explanation:

  • The chemistry reaction's balanced equation is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.

  • Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
  • The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
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2 months ago
The next 3 questions will walk you through using the Henderson-Hasselbalch equation for the following question. For each step pr
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Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 × 10^{-5}

Calculating pKb: pKb = - log (6.3 × 10^{-5})

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