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Question What is the change in pH when 10.0 cm3 of 1.00 mol dm–3 HCl is added to 1.00 dm3 of ethanoic acid / sodium ethanoate buffer containing 0.700 mol dm–3 of CH3CO2H and 0.600 mol dm–3 of CH3CO2

–? (Ka of ethanoic acid = 1.80 × 10–5 mol dm–3) Step 1: Calculate the pH of the buffer before HCl is added. Step 2: Calculate the pH of the buffer after HCl is added. Change in pH = (pH from Step 2) – (pH from Step 1)

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Step 1: To calculate the pH of the buffer before HCl is added The pH of this buffer was calculated to be 4.68 in an earlier example.

(Refer to earlier post on how to calculate the pH of a buffer: http://www.wewwchemistry.com/2012/12/example-how-to-calculate-ph-of-buffer.html)

Step 2: To calculate the pH of the buffer after HCl is added When 10.0 cm3 of 1.00 mol dm–3 HCl is added to the buffer (1.00 dm3), a reaction

takes place. The added H+ reacts with the base component of the buffer, CH3CO2

–, as follows: CH3CO2

– + H+ → CH3CO2H As a result of this reaction, [CH3CO2

–] decreases while [CH3CO2H] increases. Therefore, we need to find the new [CH3CO2H] and [CH3CO2

–] before we can proceed to use the ICE table or the Henderson-Hasselbalch equation to find the new pH of the buffer.

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To find new [CH3CO2H] and new [CH3CO2–] after HCl is added,

1 mol CH3CO2

– ≡ 1 mol H+ ≡ 1 mol CH3CO2H Moles of HCl added = (0.0100)(1.00) = 0.0100 mol Moles of CH3CO2

– left after reaction = 0.600 – 0.0100 = 0.590 mol 0.590 New [CH3CO2

–] after reaction = (1.00 + 0.0100) = 0.584 mol dm–3

Moles of CH3CO2H after reaction = 0.700 + 0.0100 = 0.710 mol

0.710 New [CH3CO2H] after reaction = (1.00 + 0.0100) = 0.703 mol dm–3

Original volume of buffer in dm3

volume of acid added in dm3

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To find the new pH of the buffer, Method 1 Constructing the ICE table,

CH3CO2H ∏ H+ + CH3CO2–

Initial conc / mol dm–3 0.703 0 0.584 Change in conc / mol dm–3 – y + y + y Eqm conc / mol dm–3 0.703 – y y 0. 584 + y ≈ 0.703 ≈ 0.584

Assumption: CH3CO2H is a weak acid dissociating in the presence of its conjugate base,

CH3CO2–, of similar concentration.

The presence of CH3CO2– suppresses the dissociation of CH3CO2H due to the

Common Ion Effect. It is thus assumed that y is insignificant, such that 0.703 – y ≈ 0.703 and 0.584 + y ≈

0.584.

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[CH3CO2

–][ H+] Ka = [CH3CO2H]

(0. 584 + y)(y) 1.80 × 10–5 = (0.703 – y)

0.584 y 1.80 × 10–5 ≈ 0.703

y = 2.17 × 10–5 ∴ The pH of buffer = –lg(2.17 × 10–5) = 4.66

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Method 2 Substituting new [CH3CO2H] and new [CH3CO2

–] directly into the Henderson-Hasselbalch equation,

[CH3CO2–] pH = pKa + lg [CH3CO2H]

(0.584 + y) pH = –lg(1.80 × 10–5) + lg (0.703 – y)

Since y is negligible (see Method 1 assumption),

0.584 pH ≈ –lg(1.80 × 10–5) + lg 0.703 = 4.66

Conclusion: pH of this buffer decreases by only 0.02 units after the addition of HCl!

This step is usually skipped.