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What is the pH of a buffer prepared by adding 0.708 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

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Answer:

a) pH before adding NaOH = 6.18

b) pH after adding NaOH = 6.45

Explanation:

The pH of a buffer is calculated using Henderson-Hasselbach equation:

  • pH = pKa + log [tex]\frac{[A^{-}]}{[HA]} \\[/tex]
  • Where pKa = -log(Ka) = -log(5.66x10⁻⁷) = 6.25

Using the given moles and volume, we can calculate the molar concentrations of the weak acid and its conjugate base:

  • [HA] = 0.708 mol / 2.00 L = 0.354 M
  • [NaA] = [A⁻] = 0.608 mol / 2.00 L = 0.304 M

So now we calculate pH:

  • pH = 6.25 + log (0.304/0.354) = 6.18

For the second question, OH⁻ moles decrease the moles of the weak acid and increase the moles of the conjugate base:

  • mol HA = Original mol HA - mol OH⁻ = 0.708 - 0.195 = 0.513 mol
  • mol A⁻ = Original mol A⁻ + mol OH⁻ = .608 + 0.195 = 0.803 mol

With the new moles, we can calculate the new molar concentrations:

  • [HA] = 0.513 mol / 2.00 L = 0.256 M
  • [A⁻] = 0.803 mol / 2.00 L = 0.402 M

Finally we calculate the new pH:

  • pH = 6.25 + log (0.402/0.256) = 6.45
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