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 2.0*10-9.

pH = 8.48

Explanation:

HOBr dissociates in water as follows:

HOBr (aq) ↔ H⁺ (aq) + OBr⁻ (aq)

With a ka expressed by:

ka = [H⁺]*[OBr⁻] / [HOBr]

We rearrange and solve for [H⁺]:

[H⁺] = ka * [HOBr] / [OBr]

Because the volume is 1 L, the moles added of HOBr and KOBr (OBr⁻) are also the molar concentration:

[H⁺] = 2.0x10⁻⁹ * 0.50 / 0.30 [H⁺] = 3.33x10⁻⁹ M

Finally we calculate the pH:

pH = - log[H⁺] pH = 8.48