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Nesterboy [21]
3 years ago
11

1.Compare and contrast the rate of solution formation between the three physical forms of salt that were placed in the vial and

not agitated with the three forms of salt that were placed in the vial and were agitated.
2. On the basis of your results, what is the relationship between the temperature of the solvent and the rate of solution formation?
3. Use your knowledge of collision theory to explain the results of your experiments in this laboratory.
Chemistry
1 answer:
lawyer [7]3 years ago
8 0

<span>The stable form of salt at standard temperature and pressure is solid. If you add water to salt, it will dissolve. However, there are certain factors that affect the rate of solution formation of salt. The rate of formation of salt solution is faster when the vial is agitated than when it is not agitated. By agitating the solution, you are increasing the surface area of the salt particles in contact with water.</span>

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A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
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Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

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pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

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5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

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As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

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1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

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0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

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As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

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