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kirill115 [55]
3 years ago
14

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 ho

br is 2.0×10−9.
Chemistry
1 answer:
Rufina [12.5K]3 years ago
7 0

Answer:

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 <u>solve for [H⁺]</u>:

  • [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 <u>calculate the pH</u>:

  • pH = -log[H⁺]
  • pH = 8.48
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