Which of the following acids should be used to prepare a buffer with a pH of 4.5? A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4 (COOH) 2, Ka = 2.9 x 10^-4C. CH3COOH, Ka = 1.8 x 10^-5D. C5H5COOH, Ka = 4.0 x 10^-6E. HBrO, Ka = 2.3 x 10^-9

Question

Kaelyn

Chemistry

14 March, 19:50

Which of the following acids should be used to prepare a buffer with a pH of 4.5? A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4 (COOH) 2, Ka = 2.9 x 10^-4C. CH3COOH, Ka = 1.8 x 10^-5D. C5H5COOH, Ka = 4.0 x 10^-6E. HBrO, Ka = 2.3 x 10^-9

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Answers (1)

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Toby Good · Answer 1

C. CH3COOH, Ka = 1.8 E-5

Explanation:

analyzing the pKa of the given acids:

∴ pKa = - Log Ka

A. pKa = - Log (1.0 E-3) = 3

B. pKa = - Log (2.9 E-4) = 3.54

C. pKa = - Log (1.8 E-5) = 4.745

D. pKa = - Log (4.0 E-6) = 5.397

E. pKa = - Log (2.3 E-9) = 8.638

We choose the (C) acid since its pKa close to the expected pH.

⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):

pH = pKa + Log ([CH3COO-]/[CH3COOH])

∴ pH = 4.5

∴ pKa = 4.745

⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])

⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])

⇒ 0.5692 = [CH3COO-]/[CH3COOH]

∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-]) / [CH3COOH]

⇒ 1.8 E-5 = [H3O+] (0.5692)

⇒ [H3O+] = 3.1623 E-5 M

⇒ pH = - Log (3.1623 E-5) = 4.5