Solution #1: A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate to 100.0 mL of water. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution #1.
ph buffer question. NEED help due tomorrow.?
2016-06-20 7:17 am
回答 (1)
2016-06-20 7:46 am
When NaOH is added to solution #1 :
CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)
Initial no. of moles of CH₃COOH = (0.400 mol) × (100.0/1000 L) = 0.0400 mol
Initial no. of moles of OH⁻ = (1.10 mol) × (55.0/1000 L) = 0.0605 mol
No. of moles of unreacted OH⁻ = (0.0605 - 0.0400) = 0.0205 mol
Final [OH⁻] = (0.0205 mol) / [(100.0 + 55.0)/1000 L] = 0.132 M
In the presence of 0.132 M OH⁻, the hydrolysis of CH₃COO⁻ and the self-dissociation of water should be neglected.
pOH = -log[OH⁻]
Final pH = 14.0 - pOH = 14.0 + log[OH⁻] = 14.0 + log(0.132) = 13.1
CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)
Initial no. of moles of CH₃COOH = (0.400 mol) × (100.0/1000 L) = 0.0400 mol
Initial no. of moles of OH⁻ = (1.10 mol) × (55.0/1000 L) = 0.0605 mol
No. of moles of unreacted OH⁻ = (0.0605 - 0.0400) = 0.0205 mol
Final [OH⁻] = (0.0205 mol) / [(100.0 + 55.0)/1000 L] = 0.132 M
In the presence of 0.132 M OH⁻, the hydrolysis of CH₃COO⁻ and the self-dissociation of water should be neglected.
pOH = -log[OH⁻]
Final pH = 14.0 - pOH = 14.0 + log[OH⁻] = 14.0 + log(0.132) = 13.1
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