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Anna007 [38]
3 years ago
8

The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butano

ic acid.
Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

Ka = 1.52 E-5

Explanation:

  • CH3-(CH2)2-COOH ↔ CH3(CH2)2COO-  + H3O+

⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]

mass balance:

⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3(CH2)2COO-]

⇒ Ka = [H3O+]²/(1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )

⇒ Ka = 1.519 E-5

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And with solution...
Tems11 [23]

Answer:

The answer to your question is: V = 6.93 L

Explanation:

Data

N₂ = 5.6 g

Volume of NH₃ = ?

                              14 g of N   ----------------  1 mol

                              5.6 g -----------------------   x

                             x = (5.6 x 1) / 14 = 0.4 mol of N

Reaction

                                N₂    +     3H₂    ⇒    2NH₃

                                1 mol of N₂   ----------------  2 moles of NH₃

                                0.4 mol of N₂ --------------   x

                               x = (0.4 x 2) / 1

                               x = 0.8 mol of NH₃

Formula

                 PV = nRT

P = 5200 torr = 6.84 atm

V = ?

n = 0.8

R = 0.082 atm L/ mol °K

T = 450°C = 723°K

Substitution

                     V = (0.8)(0.082)(723) / 6.84

                     V = 6.93 L

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3 years ago
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