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denpristay [2]
3 years ago
5

In the laboratory, a general chemistry student measured the pH of a 0.486 M aqueous solution of triethanolamine, C6H15O3N to be

10.740. Use the information she obtained to determine the Kb for this base.
Chemistry
1 answer:
STatiana [176]3 years ago
5 0

Answer:

Kb = 6.22x10⁻⁷

Explanation:

Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:

C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)

Kb is defined from concentrations in equilibrium, thus:

Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]

The equilibrium concentration of these compounds could be written as:

[C₆H₁₅O₃N] = 0.486M - X

[C₆H₁₅O₃NH⁺] = X

[OH⁻] = X

pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M

Also, Kw = [OH⁻] ₓ [H⁺];

1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]

1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]

5.495x10⁻⁴M = [OH⁻], that means <em>X = 5.495x10⁻⁴M</em>

Replacing in Kb formula:

Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]

<em>Kb = 6.22x10⁻⁷</em>

<em></em>

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What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperatu
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1) 6.0 atm.

2) 2.066 atm.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>

  • At constant T and at two different (P, and V):

<em>P₁V₁ = P₂V₂.</em>

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>

<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>

<em></em>

  • Since the container is rigid, so it has constant V.
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<em>P₁/T₁ = P₂/T₂.</em>

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>

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