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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.

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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 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

The pH of the solution  = 8.48

Chemical reaction for HOBr:

HOBr(aq) ↔ H⁺(aq) + OBr⁻(aq)

Dissociation constant can be given as:

ka = [H⁺]*[OBr⁻] / [HOBr]

Solving 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⁺] = 2.0* 10⁻⁹ * 0.50 / 0.30

[H⁺] = 3.33* 10⁻⁹ M

Calculation for  pH:

pH = -log[H⁺]

pH = 8.48

Thus, the pH of the solution is 8.48.

Find more information about pH here:

brainly.com/question/22390063

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