Question

 

PART 1

In the laboratory, a general chemistry student measured the pH of a 0.321 M aqueous solution of caffeine, C₈H₁₀N₄O₂ to be 12.045.

Use the information she obtained to determine the K_b for this base.

K_b(experiment) = _____

PART 2

In the laboratory, a general chemistry student measured the pH of a 0.551 M aqueous solution of trimethylamine, (CH₃)₃N to be 11.754.

Use the information she obtained to determine the K_b for this base.

K_b(experiment) = _____

PART 3

In the laboratory, a general chemistry student measured the pH of a 0.562 M aqueous solution of codeine, C₁₈H₂₁O₃N to be 10.834.

Use the information she obtained to determine the K_b for this base.

K_b(experiment) = _____

Answer 1

Part 1

Caffeine disssociate as

C₈H₁₀N₄O₂ + H₂O C₈H₁₁N₄O₂⁺ + OH^-

K_b = [C₈H₁₁N₄O₂⁺ ] [OH^-] / [C₈H₁₀N₄O₂ ]

pOH = 14 - pH = 14 - 12.045 = 1.955

OH^- = 10^-pOH = 10^-1.955 = 0.01109 M

According to reaction 1 mole C₈H₁₀N₄O₂ dissociate produce 1 mole of C₈H₁₁N₄O₂⁺ and 1 mole of OH^- therefore

[C₈H₁₁N₄O₂⁺ ] = [OH^-] = 0.01109 M

K_b = [0.01101][0.01109] / 0.321 = 0.000383 = 3.83 X 10^-4

K_b of Caffeine = 3.83 X 10^-4

Part 2

(CH₃)₃N disssociate as

(CH₃)₃N + H₂O (CH₃)₃NH ⁺ + OH^-

K_b = [(CH₃)₃N​H ⁺  ] [OH^-] / [(CH₃)₃N ]

pOH = 14 - pH = 14 - 11.754 = 2.246

OH^- = 10^-pOH = 10^-2.246 = 0.005675 M

According to reaction 1 mole (CH₃)₃N dissociate produce 1 mole of (CH₃)₃N​H ⁺ and 1 mole of OH^- therefore

[(CH₃)₃N​H ⁺  ] = [OH^-] = 0.005675 M

K_b = [0.005675][0.005675] / 0.551 = 0.000383 = 5.84 X 10^-5

K_b of (CH₃)₃N = 5.84 X 10^-5

Part 3

C₁₈H₂₁O₃N disssociate as

C₁₈H₂₁O₃N + H₂O C₁₈H₂₂O₃N ⁺ + OH^-

K_b = [C₁₈H₂₂O₃N ⁺ ] [OH^-] / [C₁₈H₂₁O₃N ]

pOH = 14 - pH = 14 - 10.834 = 3.166

OH^- = 10^-pOH = 10^-3.166 = 0.000682 M

According to reaction 1 mole C₁₈H₂₁O₃N dissociate produce 1 mole of C₁₈H₂₂O₃N ⁺  and 1 mole of OH^- therefore

[C₁₈H₂₂O₃N ⁺  ] = [OH^-] = 0.000682 M

K_b = [0.000682][0.000682] / 0.562 = 0.000383 = 8.28 X 10^-7

K_b of C₁₈H₂₁O₃N = 8.28 X 10^-7