Help with Part B & C please! Part A What is the pH of a buffer prepared by addin

Question

Help with Part B & C please!

Part A

What is the pH of a buffer prepared by adding 0.607 mol of the weak acid HA to 0.507 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107.

Express the pH numerically to three decimal places.

Hints

6.169

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Correct

Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the pH. The answer will be the same.

Part B

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

Express the pH numerically to three decimal places.

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Part C

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.

Express the pH numerically to three decimal places.

6.169

Explanation / Answer

A)

initially

mol of A = MV = 0.507 *2 = 1.014

mol of HA = MV = 0.607*2 = 1.214

pH = pKa + log(a-/HA)

pKa = -log(5.66*10^-7) = 6.247

pH = 6.247+ log(1.014/1.214) = 6.168

b)

mol of HCl = 0.15

after reaction

mol of A- = 1.014-0.15 = 0.864

mol of HA = 1.214+0.15 = 1.364

pH = pKa + log(a-/HA)

pH = 6.247+ log(0.864/1.364) = 6.0486

c)

mol of NaOH = 0.195

after reaction

mol of HA= 1.014-0.195= 0.819

mol of A- = 1.214+0.195= 1.409

pH = pKa + log(a-/HA)

pH = 6.247+ log(0.819/1.409) = 6.0113

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