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Pie
3 years ago
10

Consider a solution which is 0.10M in CH3COOH and 0.20M in NaCH3COO. Choose ALL of the following that are true. If Ca(CH3COO)2 i

s added, the pH becomes more basic. If a small amount of HCl is added, the pH becomes slightly more acidic. If HCl is added, the H ions react with the CH3COOH. If a small amount of NaOH is added, the pH becomes slightly more acidic. If NaOH is added, the OH- ions react with the CH3COO- ions.
Chemistry
2 answers:
Valentin [98]3 years ago
7 0

Answer:

The first 2 statements are correct. The others are false.

If Ca(CH3COO)2 is added, the pH becomes more basic.

If a small amount of HCl is added, the pH becomes slightly more acidic.

Explanation:

Step 1: Data given

Molarity of a CH3COOH solution = 0.10 M

⇒CH3COOH is a weak acid

Molarity of a NaCH3COO solution = 0.20 M

Step 2: If Ca(CH3COO)2 is added, the pH becomes more basic.

Ca(CH3COO)2 → Ca^2+ + 2CH3COO-

This means when we add Ca(CH3COO)2, the amount of CH3COO- will be increased.

CH3COO- is the conjugate base of the acid CH3COOH.

By increasing the amount of the conjugate base CH3COO- the pH will become more basic.

The first statement is true.

Step 3:  If a small amount of HCl is added, the pH becomes slightly more acidic

This is true. By adding a strong acid, the pH will become more acid.

Step 4: If HCl is added, the H ions react with the CH3COOH.

When the strong acid HCl is added. The H+ ions will react with a negative charged compound. This will not be CH3COOH but it's conjugate base CH3COO-

This statement will be false.

Step 5:  If a small amount of NaOH is added, the pH becomes slightly more acidic.

When we add HCl, ste pH will become more acid. NaOH is a strong base. When adding a strong base, the pH will become (slightly) more basic, not more acid.

This statement is false.

Step 6: If NaOH is added, the OH- ions react with the CH3COO- ions.

The OH- ions of NaOH are negative charged, they will not react with the CH3COO- ions because those are also negative charged.

This statement is false?

The first 2 statements are correct. The others are false.

If Ca(CH3COO)2 is added, the pH becomes more basic.

If a small amount of HCl is added, the pH becomes slightly more acidic.

Ahat [919]3 years ago
5 0

Answer:

If Ca(CH₃COO)₂ is added, the pH becomes more basic.

If a small amount of HCl is added, the pH becomes slightly more acidic.

Explanation:

In the buffer:

CH₃COOH (weak acid) and NaCH₃COO (conjugate base), the equilibrium in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

CH₃COOH will react if a base is added and CH₃COO⁻ reacts if an acid is added decreasing the pH change. Thus:

If Ca(CH₃COO)₂ is added, the pH becomes more basic. <em>TRUE. </em>Here, you are adding CH₃COO⁻ (Conjugate base), as a base, pH will decreases.

If a small amount of HCl is added, the pH becomes slightly more acidic. <em>TRUE. </em>As buffer, the addition of strong acids as HCl decrese slightly the pH of the solution.

If HCl is added, the H ions react with the CH₃COOH. <em>FALSE. </em>The strong acid reacts with the conjugate base, CH₃COO⁻

If a small amount of NaOH is added, the pH becomes slightly more acidic. <em>FALSE. </em>The addition of a strong base as NaOH become pH slightly more basic.

If NaOH is added, the OH⁻ ions react with the CH₃COO⁻ ions. <em>FALSE. </em>A strong acid reacts with the weak acid, CH₃COOH.

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Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

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Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

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Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

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In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

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