Aqueous Ionic Equilibrium?

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2021-03-17 10:25 am

A 1.0-L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution upon addition of 88.0 mL of 1.00 M HCl to the original buffer.

回答 (1)

Uncle Michael

2021-03-17 11:44 am

✔ 最佳答案

Moles of HCl added = (1.00 mol/L) × (88.0/1000 L) = 0.088 mol

Each mole of HCl added reacts with 1 mol C₂H₃O₂⁻ to form 1 mole of HC₂H₃O₂.
Moles of C₂H₃O₂⁻ after addition = (0.100 - 0.088) mol = 0.012 mol
Moles of HC₂H₃O₂ after addition = (0.100 + 0.088) mol = 0.188 mol

After addition of HCl:
[C₂H₃O₂⁻]/[HC₂H₃O₂] = (Moles of C₂H₃O₂⁻)/(Moles of HC₂H₃O₂) = 0.012/0.188

Consider the dissociation of HC₂H₃O₂:
HC₂H₃O₂(aq) + H₂O(ℓ) ⇌ C₂H₃O₂⁻(aq) + H₃O⁺(aq) Ka = 1.8 × 10⁻⁵

Henderson-Hasselbalch equation:
pH = pKa + log([C₂H₃O₂⁻]/[HC₂H₃O₂])
pH = -log(1.8 × 10⁻⁵) + log(0.012/0.188)
pH = 3.55

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