中和滴定曲線 (5回目)

中和滴定曲線 (5回目)

2026/1/31(土)

中和滴定曲線 (5回目)

(neutralization titration)

2価の弱酸と1価の強塩基の塩と1価の強酸の滴定曲線
[Na₂CO₃aqをHClaqで滴定]

炭酸ナトリウム水溶液と塩酸の反応式
Na₂CO₃ + HCl → NaCl + NaHCO₃
NaHCO₃ + HCl → NaCl + H₂CO₃ (CO₂ + H₂O)

c,c',V,V'は元の酸･塩基のモル濃度と体積
Ca = cV/(V+V') , Cb = c'V'/(V+V') は滴定中の酸･塩基のモル濃度

水溶液中に[H₂A],[HA^-],[H⁺],[A^2-],[B⁺],[OH^-]が存在する
Ca = [A^2-] + [H₂A] + [HA^-] , Cb = [B⁺]  … 酸･強塩基のモル濃度
Kw = [H⁺][OH^-]                         … 水のイオン積
Ka1 = [H⁺][HA^-]/[H₂A]                  … 酸の電離定数(第1)
Ka2 = [H⁺][A^2-]/[HA^-]                  … 酸の電離定数(第2)
[B⁺] + [H⁺] = 2[A^2-] + [HA^-] + [OH^-]     … 電気的中性

[H⁺]⁴ + (Ka1 + Cb)[H⁺]³ + {Ka1(Ka2 + Cb - Ca) - Kw}[H⁺]²
+ Ka1{Ka2(Cb - 2Ca) - Kw}[H⁺] - Ka1Ka2Kw = 0

元の酸･塩基で表すと
[H⁺]⁴ + {Ka1 + c'V'/(V+V')}[H⁺]³ + {Ka1{Ka2 + (cV-c'V')/(V+V')}
- Kw}[H⁺]² + Ka1{(cV-2c'V')/(V+V')Ka2 - Kw}[H⁺] - Ka1Ka2Kw = 0

A = [H⁺]{[H⁺]{[H⁺]([H⁺] + Ka1) + Ka1(Ka2 - Kw)} - Ka1Kw} - Ka1Ka2Kw
B = [H⁺]cKa1([H⁺] + 2Ka2) - A
C = c'[H⁺]{[H⁺]([H⁺] + Ka1) + Ka1Ka2} + A
V' = (B/C)V

上記式の[B⁺]を[B⁺]-[C^-]に置き換える

c,c',V,V'は元の塩(B₂A)と元の強酸酸(HC)のモル濃度と体積
Ca = cV/(V+V'), Cb = (2cV-c'V')/(V+V')は滴定中の酸と塩基のモル濃度

水溶液中に[H₂A],[HA^-],[H⁺],[A^2-],[B⁺],[OH^-],[C^-]が存在する
Ca = [A^2-] + [H₂A] + [HA^-] , Cb = [B⁺]-[C^-] … 酸･強塩基のモル濃度(塩)
Kw = [H⁺][OH^-]                              … 水のイオン積
Ka1 = [H⁺][HA^-]/[H₂A]                       … 酸の電離定数(第1)
Ka2 = [H⁺][A^2-]/[HA^-]                       … 酸の電離定数(第2)
([B⁺]-[C^-]) + [H⁺] = 2[A^2-] + [HA^-] + [OH^-]   … 電気的中性

[H⁺]⁴ + (Ka1 + Cb)[H⁺]³ + {Ka1(Ka2 + Cb - Ca) - Kw}[H⁺]²
+ Ka1{Ka2(Cb - 2Ca) - Kw}[H⁺] - Ka1Ka2Kw = 0

元の塩と酸で表すと
[H⁺]⁴ + {Ka1 + (2cV-c'V')/(V+V')}[H⁺]³ + {Ka1{Ka2 + (cV-c'V')/(V+V')}
- Kw}[H⁺]² + Ka1{c'V'/(V+V')Ka2 - Kw}[H⁺] - Ka1Ka2Kw = 0

V'を求める式
(V+V')[H⁺]⁴ + {(V+V')Ka1 + 2cV-c'V'}[H⁺]³
+ {Ka1((V+V')Ka2 + (cV-c'V')) - (V+V')Kw}[H⁺]²
- Ka1{Ka2c'V' + (V+V')Kw}[H⁺] - (V+V')Ka1Ka2Kw = 0

A = [H⁺]⁴ + Ka1[H⁺]³ + Ka1(Ka2 - Kw)[H⁺]² - Ka1Kw[H⁺] - Ka1Ka2Kw
= [H⁺]{[H⁺]³ + Ka1[H⁺]² + Ka1(Ka2 - Kw)[H⁺] - Ka1Kw} - Ka1Ka2Kw
= [H⁺]{[H⁺]{[H⁺]² + Ka1[H⁺] + Ka1(Ka2 - Kw)} - Ka1Kw} - Ka1Ka2Kw
= [H⁺]{[H⁺]{[H⁺]([H⁺] + Ka1) + Ka1(Ka2 - Kw)} - Ka1Kw} - Ka1Ka2Kw
B = -V(2c[H⁺]³ + cKa1[H⁺]² + A)
= -V{c[H⁺]²(2[H⁺] + Ka1) + A}
C = -c'[H⁺]³ - c'Ka1[H⁺]² - c'Ka1Ka2[H⁺] + A
= -[H⁺]c'{[H⁺]² + Ka1[H⁺] + Ka1Ka2} + A
= -[H⁺]c'{[H⁺]([H⁺] + Ka1) + Ka1Ka2} + A

A = [H⁺]{[H⁺]{[H⁺]([H⁺] + Ka1) + Ka1(Ka2 - Kw)} - Ka1Kw} - Ka1Ka2Kw
B = c[H⁺]²(2[H⁺] + Ka1) + A
C = c'[H⁺]{[H⁺]([H⁺] + Ka1) + Ka1Ka2} - A
V' = (B/C)V

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