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Aspirin(acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. Todetermine its acid-dissociation constant, a student dissolved 2.00g of aspirin in 0.600 Lof water and measured the pH. What wasthe

Ka valuecalculated by the student if the pH of the solution was2.61?

Part B

A 0.100 Msolution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate theKb for ethylamine.

Expressyour answer numerically using two significant figures.

Respuesta :

Answer:

A) Ka = 3.7 * 10^-4

B) Kb = 5.9 * 10^-4

Explanation:

Step 1: Data given

Mass of aspirin = 2.00 grams

Volume of water = 0.600 L

pH = 2.61

Step 2: Calculate moles of acetylsalicylic acid

Moles acetylsalicylic acid = mass acetylsalicylic acid / molar mass acetylsalicylic acid

Moles acetylsalicylic acid = 2.00 grams / 180.16 g/mol

Moles acetylsalicylic acid = 0.0111 moles

Step 3: Calculate molarity of acetylsalicylic acid

Molarity acetylsalicylic acid = 0.0111 moles / 0.600 L

Molarity acetylsalicylic acid = 0.0185 M

Step 4: initial concentration

[C9H8O4] = 0.0185 M

[H3O+] = 0M

[C9H7O4-] = 0M

Step 5: Calculate the concentration at equilibrium

[C9H8O4] = 0.0185 -x M

[H3O+] = xM

[C9H7O4-] = xM

Step 6: Calculate Ka

Ka = [H3O+][C9H7O4-] / [C9H8O4]

Ka = x² / (0.0185 - x)

pH = 2.61 ⇒ [H3O+] = 10^-pH = 10^-2.61 = 0.00245 = x

Ka = (0.00245)² / (0.0185 - 0.00245) = 3.7 * 10^-4

Part B

A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate theKb for ethylamine.

C2H5NH2 + H2O ⇒ C2H5NH3+ + OH-

The initial molarity

[C2H5NH2] = 0.100 M

[C2H5NH3+] = 0M

[OH-] = 0M

The molarity at equilibrium

[C2H5NH2] = 0.100 -x M

[C2H5NH3+] = xM

[OH-] = xM

pH = 11.87

pOH = 14.00 - pH = 14.00 - 11.87 = 2.13

[OH-] = 10^-pOH = 10^-2.13 = 0.0074 = x

Kb = [C2H5NH3+][OH-] / [C2H5NH2] = x² / (0.100 - x) = (0.0074)² / (0.100 - 0.0074) = 5.9 * 10^-4