Calculate the pH of the solution made by adding 0.50 mol of HOBr and 0.30 mol of KOBr to 1.00 L of water. The value of Ka for HOBr is 2.0Ă—10â’9. Express your answer numerically using two decimal places.

Respuesta :

Here we have to get the the [tex]P^{H}[/tex] of the solution containing HOBr and KOBr.

The [tex]P^{H}[/tex] of the solution is 8.47

The mixture of HOBr and KOBr is a buffer solution. The [tex]P^{H}[/tex] of the buffer solution is determined by the Henderson equation which is:

[tex]P^{H}[/tex] = [tex]P_{K}_{a}[/tex] + log [tex]\frac{[salt]}{[acid]}[/tex]

Here [salt] i.e. KOBr is 0.30 mol/L and [Acid] is 0.50 mol/L.

The [tex]K_{a}[/tex] value of HOBr is 2×10⁻⁹.

We know, [tex]P_{K}_{a}[/tex] = -log[tex]K_{a}[/tex]

Or, [tex]P_{K}_{a}[/tex] = -log (2×10⁻⁹)

Or, [tex]P_{K}_{a}[/tex] = 8.698

On plugging the values:

[tex]P^{H}[/tex] = 8.698 + log [tex]\frac{0.3}{0.5}[/tex]

Or, [tex]P^{H}[/tex] = 8.698 - 0.221

Or, [tex]P^{H}[/tex] = 8.47

Thus the [tex]P^{H}[/tex] of the solution is 8.47.