Question

A beaker with 145 mLmL of an acetic acid buffer with a pHpH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 mol L−1mol L−1. A student adds 6.70 mLmL of a 0.360 mol L−1mol L−1 HClHCl solution to the beaker. How much will the pHpH change? The pKapKa of acetic acid is 4.760.

Express your answer numerically to two decimal places. Use a minus ( −− ) sign if the pHpH has decreased.

Answer 1

From Henderson Hasselbalch equation:

pH = pKa + log (B/A)

B : Base

A : Acid

5.000 = 4.760 + log (B/A)

B/A = 10^5-4.76

B/A = 1.7378

B = 1.7378 A

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Also, given that:

A + B = (145 mL) x (0.1 M)

A + 1.7378 A = 14.5 mmol

A = 5.29622 mmol

B = 14.5 - 5.29622 mmol

B = 9.20378 mmol

No.of moles of acetate (B) = 9.20378 mmol

No.of moles of acetic acid (A) = 5.29622 mmol

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No.of moles of HCl added = (6.70 mL) x (0.360 M) = 2.412 mmol

Final no.of moles of acetate (B) = 9.20378 mmol - 2.412 mmol = 6.791778358 m.mol

Final No.of moles of acetic acid (A) = 5.29622 mmol +  2.412 mmol = 7.708221642 m.mol

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Now, Applying Henderson Hasselbalch equation for the new buffer:

pH = pka + log (B/A)

pH = 4.760 + log (6.791778358 / 7.708221642)

pH = 4.70503

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Hence, Change in pH = 4.70503 - 5.000

Change in pH = -0.295

• Answer : Change in pH = - 0.295