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
`([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
`(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.