Calculate the molar concentration of H+ ion of a solution that is 5.30×10-2 M in C6H5CO2H and 2.94×10-2 M in the salt C6H5CO2Li.
Ka = 0.000063
C6H5CO2H(aq) = C6H5CO2-(aq) + H+(aq)
Calculate molar concentration of H+?
2016-05-04 4:18 pm
回答 (2)
2016-05-04 5:08 pm
✔ 最佳答案
C₆H₅CO₂H(aq) ⇌ C₆H₅CO₂⁻(aq) + H⁺(aq) ... Kₐ = 0.000063Due to small Kₐ and the common ion effect :
[C₆H₅CO₂H] ≈ [C₆H₅CO₂H]ₒ = 5.30 × 10⁻² M
[C₆H₅CO₂⁻] ≈ [C₆H₅CO₂⁻]ₒ = 2.94 × 10⁻² M
Kₐ = [C₆H₅CO₂⁻] [H⁺] / [C₆H₅CO₂H]
Molar concentration of H⁺ ion, [H⁺]
= Kₐ [C₆H₅CO₂H] / [C₆H₅CO₂⁻]
= 0.000063 × (5.30 × 10⁻²) / (2.94 × 10⁻²) M
= 1.14 × 10⁻⁴ M
2016-05-04 4:27 pm
Calculate pH using Henderson - Hasselbalch equation:
pKa = -log 0.000063 = 4.20
pH = pKa + log ([salt]/[acid] )
pH = 4.20 + log (2.94*10^-2) / (5.3*10^-2)
pH = 4.20 + log 0.5547
pH = 4.20 + (-0.26)
pH = 3.94
[H+] = 10^-pH
[H+] = 10^-3.94
[H+] = 1.15*10^-4M
pKa = -log 0.000063 = 4.20
pH = pKa + log ([salt]/[acid] )
pH = 4.20 + log (2.94*10^-2) / (5.3*10^-2)
pH = 4.20 + log 0.5547
pH = 4.20 + (-0.26)
pH = 3.94
[H+] = 10^-pH
[H+] = 10^-3.94
[H+] = 1.15*10^-4M
收錄日期: 2021-04-18 14:46:54
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