Question

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.

Answer 1

Here we have to get the the $P^{H}$ of the solution containing HOBr and KOBr.

The $P^{H}$ of the solution is 8.47

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

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

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

The $K_{a}$ value of HOBr is 2×10⁻⁹.

We know, $P_{K}_{a}$ = -log$K_{a}$

Or, $P_{K}_{a}$ = -log (2×10⁻⁹)

Or, $P_{K}_{a}$ = 8.698

On plugging the values:

$P^{H}$ = 8.698 + log $\frac{0.3}{0.5}$

Or, $P^{H}$ = 8.698 - 0.221

Or, $P^{H}$ = 8.47

Thus the $P^{H}$ of the solution is 8.47.