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3 years ago

Identify the acid, base, conjugate acid and conjugate base in the following reactions:

Chemistry

1 answer:

Leni [432]3 years ago

4 0

Answer:

a. NH₃ : base

CH₃COOH (acetic acid) : acid

NH₄⁺ : conjugate acid

CH₃COO⁻ : conjugate base

b. HClO₄ (perchloric acid) : acid

NH₃ : base

ClO₄⁻ : conjugate base

NH₄⁺ : conjugate acid

Hope this helps.

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Sulfuric acid reacts with aluminum hydroxide by double replacement. if 34 g of sulfuric acid react with 33 g of aluminum hydroxi

Lina20 [59]

Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.

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3 years ago

Is this equation balanced or unbalanced? C+02=CO2

LenKa [72]

This equation is balanced

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3 years ago

I know the photo trash but i need answers for the top middle and last question please :)

Firlakuza [10]

Answer:

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2 years ago

A 1.00 liter solution contains 0.43 M hydrofluoric acid and 0.56 M potassium fluoride. If 0.280 moles of potassium hydroxide are

kolbaska11 [484]

Answer:

Answers are in the explanation

Explanation:

Equlibrium of HF in H₂O is:

HF + H₂O ⇄ F⁻ + H₃O⁺

Now, the KOH reacts with HF, thus:

KOH + HF → F⁻ + H₂O

That means after reaction, concentration of HF decrease increasing F⁻ concentration.

Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.

For the statements:

A. The number of moles of HF will increase. FALSE. HF react with KOH, thus, moles of HF decrease

B. The number of moles of F- will decrease. FALSE. The reaction produce F⁻ increasing its moles.

C. The equilibrium concentration of H₃O⁺ will increase. FALSE. The equilibrium shift to the left decreasing concentration of H₃O⁺

D. The pH will decrease. FALSE. As the H₃O⁺ concentration decrease, pH will increase

E. The ratio of [HF] / [F-] will remain the same. FALSE. Because moles of HF are decreasing whereas F- moles are increasing changing, thus, ratio.

6 0

3 years ago

Rank these acids according to their expected pKa values.

givi [52]

Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO- + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

3 0

3 years ago