Question

1. For each of the weak acids listed below, consider how you would make a buffer solution where the weak acid has a concentration of 0.5 M at equilibrium.
   1. Determine what molarity for each potassium conjugate base salt would need to be added to make 500 mL of an effective buffer with a pH of 5.32.
   2. Which acid would make the best buffer at this pH?
   3. Which conjugate base salt would be the most economical (cheapest) out of the three choices?   

molar masses (g/mol): C, 12.01; H, 1.008; O, 16.00; K, 39.1; N, 14.01.

acid	Ka	Potassium conjugate base salt	Cost of potassium conjugate base salt
CH3COOH	1.8 × 10–5		$0.32/gram
H2CO3	4.2 × 10–7		$0.21/gram
HNO2	4.6 × 10–4		$0.94/gram

Answer 1

First we calculate the pKa of each acid, with the equation:

pKa = -log Ka

CH3COOH: Ka = 4.74

H2CO3: Ka = 6.38

HNO2: Ka = 3.34

a) We calculate, by means of the Henderson Hasselbach equation cleared, the required salt concentration:

[Salt] = [Acid] * 10 ^ (pH-pKa)

CH3COOH: 1.90 M

H2CO3: 0.044 M

HNO2: 47.75 M

b) The best buffer is the one with the pKa closest to the pH needed, in this case: CH3COOH.

c) Calculate the price of each salt to form the buffer, with the equation:

$ Sal = [Salt] * 0.5 L * Cost * MM

CH3COOK = 1.9 * 0.5 * 0.32 * 98.15 = 29.83 $

KHCO3 = 0.044 * 0.5 * 0.21 * 100.12 = 0.463 $

KNO2 = 47.75 * 0.5 * 0.94 * 85.1 = 1909.86 $

We can see that the most economical is KHCO3.

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