Determine the pH during the titration of 27.1 mL of 0.139 M hydroiodic acid by 0.131 M potassium hydroxide at the following points:

1. Before the addition of any potassium hydroxide.
2. After the addition of 14.4 mL of potassium hydroxide.
3. At the equivalence point.
4. After adding 35.9 mL of potassium hydroxide.

In titration, the pH during Titration changes based on the amount of titrant added, starting from the initial pH of the acid, decreasing when some titrant is added (but not enough to neutralize), becoming neutral at the equivalence point, and then becoming basic when excess titrant is added.

Explanation:

The pH during the titration process can be determined as follows:

Before the addition of any potassium hydroxide, the solution only contains hydroiodic acid. The pH can be calculated using the formula pH = -log[H+], where [H+] is the concentration of hydronium ions. Since it is a strong acid, the concentration of H+ ions is the same as the concentration of the acid, thus pH = -log(0.139) = 0.86.

After adding 14.4 ml of 0.131 M KOH, the moles of KOH added can be calculated by multiplying its volume and molarity. Here, we have (14.4/1000)*0.131=0.00189 mol. Since this is less than the initial amount of acid, it fully reacts with the acid, and some acid remains. The remaining concentration of HI is then (0.139* (27.1-14.4)/27.1) M. The pH is then -log of this number.

At the equivalence point, all the acid has reacted with the base. Since HI is a strong acid and KOH is a strong base, the solution only has water and the salt, which doesn't hydrolyze. This implies a neutral pH of 7.

After adding 35.9 ml of KOH, the solution now has excess KOH, making the solution basic. The extra OH- can be calculated in a similar fashion to (2). Assuming complete dissociation, the pOH is then -log[OH-], and the pH can be found by subtracting pOH from 14.

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