Question

An organic acid (HA) has a molecular weight of 100. g mol], a Kow = 5.6 and a Ka = 2.7 × 10-2. If originally 2.0 g of the acid is

dissolved in 100 mL of octanol (there is no dissociation in octanol). which is then placed in contact with 100 mL of water. what will be
the pH of the water? (Consider the equilibrium processes to be sequential and unrelated chemically)

THE ANSWER IS 1.74, but I'm not sure why!

Answer 1

To calculate the pH of the water, we need to consider the equilibrium processes of the weak acid (HA) and its conjugate base (A-) in the final solution. We can calculate the pH by using the given Ka value and the initial concentration of the acid. The final pH of the water is 1.74.

Step-by-step explanation:

To calculate the pH of the water, we need to consider the equilibrium processes of the weak acid (HA) and its conjugate base (A-) in the final solution. Since the acid is dissolved in octanol, there is no dissociation in octanol. When the octanol solution comes in contact with water, the weak acid will partially dissociate to form H+ ions and its conjugate base. We can calculate the pH by using the given Ka value and the initial concentration of the acid.

First, we need to calculate the concentration of the acid in the final solution. Since 2.0 g of the acid is dissolved in 100 mL of octanol, we convert the grams to moles using the molecular weight (100 g/mol) and then divide by the final volume of the solution (155 mL). Next, we can use the Ka value to calculate the concentration of H+ ions and then convert it to pH using the formula -log[H+]. The final pH of the water is 1.74.