Answer :

The pH of the 0.35M KOBr solution is approximately 4.88. The correct option is not given, so, none of these is correct.

To find the pH of the solution, we need to calculate the concentration of H+ ions produced by the dissociation of HOBr. Since KOBr is a salt of a weak acid (HOBr), we can use the expression for the equilibrium constant (Kₐ) to find the concentration of H+ ions.

Kₐ = [H+][OBr-]/[HOBr]

Given Kₐ = 2.5 × 10⁻⁹, and assuming x as the concentration of H+ ions, we can set up the equilibrium expression:

2.5 × 10⁻⁹ = x² / (0.35 - x)

As KOBr is a salt, it will dissociate completely, leading to the formation of the same concentration of OBr- ions as the concentration of KOBr. Hence, we can simplify the equation:

2.5 × 10⁻⁹ = x² / 0.35

Solving for x gives the concentration of H+ ions.

x = √(2.5 × 10⁻⁹ × 0.35)

x ≈ 1.33 × 10⁻⁵ M

To find the pH, we use the formula:

pH = -log[H+]

pH ≈ -log(1.33 × 10⁻⁵)

pH ≈ 4.88

Since we are dealing with a weak acid, the pH will be slightly greater than the pKₐ. Therefore, the final pH of the solution is approximately 4.88. The correct option is not given, so, none of these is correct.

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