Question

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

Answer 1

The acid with a pKa value closest to the desired pH of 4.5 is acetic acid (CH₃COOH), with a pKa of 4.74, making it the best candidate for preparing the buffer.

Step-by-step explanation:

To determine the suitable acid for preparing a buffer with a pH of 4.5, we need to find the acid whose pKa value is closest to the desired pH since the buffer's pH is most effective when it's near the pKa of the buffering acid. The pKa is calculated by taking the negative logarithm (base 10) of the Ka value.

To decide which acid to use:

HOC₆H₄OCOOH, Ka = 1.0 x 10-3, pKa = 3

C₆H₄(COOH)₂, Ka = 2.9 x 10-4, pKa = 3.54

CH₃COOH, Ka = 1.8 x 10-5, pKa = 4.74

C₅H₅COOH, Ka = 4.0 x 10-6, pKa = 5.40

HBrO, Ka = 2.3 x 10-9, pKa = 8.64

Looking at these values, the acid with the pKa value closest to 4.5 is CH₃COOH (acetic acid), with a pKa of 4.74. This makes acetic acid the best candidate for preparing a buffer at pH 4.5.

Answer 2

C. CH3COOH, Ka = 1.8 E-5

Step-by-step 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