Question

1. Lactic acid, HC₃H₅O₃, is a substance found in sour milk products such as yogurt, and is produced naturally by fermentation in body cells during normal metabolism and exercise The Ka for lactic acid is 38 x 10^-4. Its sodium salt, sodium lactate, NaC₃H₅O₃, is used in foods as a preservative and acidity regulator. It is also used in shampoo products as an effective humectant and moisturizer. Determine the pH of a solution prepared by adding 4.5 grams of sodium lactate and 2.0 grams of lactic acid to 500 ml of total solution.

2. What is the new pH of a solution if 5.0 ml of 0.10 M HBr is added to 200 ml of a buffer system that is      0.10 M Benzoic acid and 0.10 M sodium benzoate. (The Ka of benzoic acid is 6.3 x 10^-5)

3. Propionic acid, C₂H₅COOH, is an oily liquid that has a slightly pungent, disagreeable rancid odor. It is found in dairy products in small amounts. Its sodium salt, sodium propionate, C₂H₅COONa, has wide application as a food preservative, fungicide and mold inhibitor. What is the ratio of propionic acid to sodium propionate in a buffer solution that has a pH of 4.26, if the Ka of propionic acid is 1.38 x 10^-5? If exactly 1.0 liters of the buffer solution is to be prepared, how many grams of C₂H₅COONa will be required?

Answer 1

Answer:

1. Here, we will use the Henderson Equation for the calculation of pH.

Molar mass of lactic acid = 90.08g/mol ; Molar mass of sodium lactate = 112.06 g/mol

Given K_a value of lactic acid = 38 x 10^-4 ; pKa = 2.42

Moles of acid = 2.0/90.08 = 0.0222 ; Moles of base = 0.0401

Molarity of acid = 0.0222/0.5 = 0.0444 ; Molarity of base = 0.0802

pH = pKa + log (base/acid) = 2.42 + log (0.0802/0.0444) = 2.67

2. 200 mL buffer contains =  0.02 moles of benzoic acid and 0.02 moles of sodium benzoate

5.0 mL of 0.10 HBr contains = 5x10^-4 moles of HBr.

As HBr is strong acid. Total volume of Solution = 205 mL

Now the moles of benzoic acid are 0.0195 and sodium benzoate are 0.0205 at the equilibrium.

[Acid] = 0.0951 M and [Base] = 0.1000 M

pH = pKa + log(Base/Acid) = 4.20 + 0.022 = 4.222

3. pH of buffer = 4.26

pKa = 4.86 (K_a = 1.38 x 10^-5)

log(base/acid) = 4.26 - 4.86 = -0.6

Base/acid = 0.2511

Base = 0.2511Acid

Molar mass of acid = 74.08 g/mol; Molar mass of sodium salt = 96.07 g/mol

Depending upon the concentration of common ion, we can take any amount of salt.

E.g [0.1 M] total is the total concentration of ions (say acid + base)

C₂H₅COONa + C₂H₅COOH = 0.1 moles (or 0.1 M)

and C₂H₅COONa = 0.2511 x C₂H₅COOH

C₂H₅COOH + 0.2511 C₂H₅COOH = 0.1 or acid = 0.0799 moles or 0.0799 M

C₂H₅COONa = 0.1 - 0.0799 = 0.0201 moles or 0.0201 M

pH = 4.86 + log(0.0201/0.0799) = 4.26 (PROVED)

Grams of C₂H₅COONa required = 0.0201 mol x 96.07 g/mol = 1.931 g

Thanks