Gen Chem pH Calculations?

All of the three questions are on calculations of buffer solutions. To find the pH, the dissociation constant (Ka or Kb) and the concentrations of the weak acid/base and its conjugate base/acid should be firstly found.


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a)
Molar mass of NaC₂H₃O₂ = (23.0 + 12.0×2 + 1.0×3 + 16×2) g/mol = 82 g/mol
No. of moles of NaC₂H₃O₂ = (25.5 g) / (82 g/mol) = 0.311 mol
[NaC₂H₃O₂] = [(25.5 g) / (82 g/mol)] / (500/1000 L) = 0.622 M

NaC₂H₃O₂ completely dissociates in water. Hence, [C₂H₃O₂⁻] = 0.622 M

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

pH = -log(Ka) - log([HC₂H₃O₂]/[C₂H₃O₂⁻]) = -log(1.8 × 10⁻⁵) - log(0.550/0.622) = 4.80


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b)
When HCl is added to NH₃ solution :
NH₃(aq) + H⁺(aq) → NH₄⁺(aq)
After reaction :
[NH₃] = (0.10 × 50.00 - 0.10 × 30.00) / (50.00 + 30.00) = 0.0250 M
[NH₄⁺] = (0.1 × 30.00) / (50.00 + 30.00) = 0.0375 M

Consider the dissociates of NH₃ :
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) ...... Kb = 1.8 × 10⁻⁵

pOH = -log(Kb) - log([NH₃]/[ NH₄⁺]) = -log(1.8 × 10⁻⁵) - log(0.0250/0.0375) = 4.92
pH = p(Kw) - pOH = 14.00 - 4.92 = 9.08


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c)
When NaOH is added to HC₂H₃O₂ :
HC₂H₃O₂(aq) + OH⁻(aq) → C₂H₃O₂⁻(aq) + H₂O(l)
[HC₂H₃O₂] = [0.250 - (0.0060/0.300)] = 0.230 M
[C₂H₃O₂⁻] = [0.560 + (0.0060/0.300)] = 0.580 M

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

pH = -log(Ka) - log([HC₂H₃O₂]/[C₂H₃O₂⁻]) = -log(1.8 × 10⁻⁵) - log(0.230/0.580) = 5.15