中和滴定曲線 (3回目)

中和滴定曲線 (3回目)

2026/1/17(土)

中和滴定曲線 (3回目)

(neutralization titration)

2価の酸と1価の強塩基の滴定曲線
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^-],[A^2-],[H⁺],[B⁺],[OH^-]が存在する
Ca = [A^2-] + [HA^-] + [H₂A] , Cb = [B⁺] + [BOH]… 酸･塩基のモル濃度
Kw = [H⁺][OH^-]                               … 水のイオン積
Ka1 = [H⁺][HA^-]/[H₂A] , Ka2 = [H⁺][A^2-]/[HA^-]   … 酸の電離定数(第1,2)
[B⁺] + [H⁺] = 2[A^2-] + [HA^-] + [OH^-]           … 電気的中性
Ka = Ka1Ka2 = [H⁺]²[A^2-]/[H₂A]

中和滴定曲線 (1回目)
より
[A^n-] = CaKa/Σ_j=0ⁿ([H⁺]^jΠ_i=1^n-jKa_i)
Caα_k = [H_n-_kA^k^-] = [H⁺]^n-k[A^n-]/(Π_i=k+1ⁿKa_i)
x = Σ(k[H_n-kA^k-]) = Σ{k[H⁺]^n-k(Π_i=2^kKa_i)}[A^n-]/(Π_i=2ⁿKa_i)
[B^m+] = CbKb/Σ_j=0^m([OH^-]^jΠ_i=1^m-jKb_i)
Cbβ_k = [B(OH)_m-k^k+] = [OH^-]^m-k[B^m+]/(Π_i=k+1^mKb_i)
y = Σ(k[B(OH)_m-k^k+]) = Σ{k[OH^-]^m-k(Π_i=2^kKb_i)}[B^m+]/(Π_i=2^mKb_i)
[OH^-] = Kw/[H⁺]
[H⁺]² - (x - y)[H⁺] - Kw = 0

n = 2
[A^n-] = CaKa/Σ_j=0ⁿ([H⁺]^jΠ_i=1^n-jKa_i)より
[A^2-] = CaKa/Σ_j=0²([H⁺]^jΠ_i=1^2-jKa_i)
= CaKa/([H⁺]⁰Π_i=1^2-0Ka_i+[H⁺]¹Π_i=1^2-1Ka_i+[H⁺]²Π_i=1^2-2Ka_i)
= CaKa/(Ka₁Ka₂+[H⁺]Ka₁+[H⁺]²) = CaKa/(Ka+[H⁺]Ka₁+[H⁺]²)

Caα_k = [H_n-_kA^k^-] = [H⁺]^n-k[A^n-]/(Π_i=k+1ⁿKa_i)より
Caα₁ = [HA^-] = [H⁺]^2-1[A^2-]/(Π_i=1+1²Ka_i) = [H⁺][A^2-]/Ka₂
Caα₂ = [A^2-] = CaKa/(Ka+[H⁺]Ka₁+[H⁺]²)
x = Σ(k[H_n-kA^k-]) = [HA^-] + 2[A^2-] = [H⁺][A^2-]/Ka₂ + 2[A^2-]
= ([H⁺]+2Ka₂)[A^2-]/Ka₂

Cbβ = [B⁺] = Cb
y = Σ(k[B(OH)_m-k^k+]) = [B⁺] = Cb

-(x-y)[H⁺] = -{([H⁺]+2Ka₂)[A^2-]/Ka₂ - Cb}[H⁺]
= [H⁺]Cb - ([H⁺]²+2[H⁺]Ka₂)[A^2-]/Ka₂
を
[H⁺]² - (x - y)[H⁺] - Kw = 0
に代入
[H⁺]² + [H⁺]Cb - ([H⁺]²+2[H⁺]Ka₂)[A^2-]/Ka₂ - Kw = 0
に
[A^2-]/Ka₂ = Ca(Ka/Ka₂)/(Ka+[H⁺]Ka₁+[H⁺]²) = CaKa₁/(Ka+[H⁺]Ka₁+[H⁺]²)
を代入
(Ka+[H⁺]Ka₁+[H⁺]²)[H⁺]² + (Ka+[H⁺]Ka₁+[H⁺]²)[H⁺]Cb
- ([H⁺]²+2[H⁺]Ka₂)CaKa₁ - (Ka+[H⁺]Ka₁+[H⁺]²)Kw
= [H⁺]²Ka+[H⁺]³Ka₁+[H⁺]⁴+[H⁺]CbKa+[H⁺]²CbKa₁+[H⁺]³Cb
- [H⁺]²CaKa₁-2[H⁺]CaKa-KaKw-[H⁺]Ka₁Kw-[H⁺]²Kw
= [H⁺]⁴+(Ka₁+Cb)[H⁺]³+{Ka-Kw+Ka₁(Cb-Ca)}[H⁺]²
+ {Ka(Cb-2Ca)-Ka₁Kw}[H⁺]-KaKw
= [H⁺]⁴+(Ka₁+Cb)[H⁺]³+{Ka₁(Ka₂+Cb-Ca)-Kw}[H⁺]²
+ Ka₁{Ka₂(Cb-2Ca)-Kw}[H⁺]-KaKw
[H⁺]⁴+(Ka₁+Cb)[H⁺]³+{Ka₁(Ka₂+Cb-Ca)-Kw}[H⁺]²
+ Ka₁{Ka₂(Cb-2Ca)-Kw}[H⁺]-KaKw = 0

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

Ca = cV/(V+V') , Cb = c'V'/(V+V')

Caα₁ = [HA^-] = [H⁺][A^2-]/Ka2
Caα₂ = [A^2-] = CaKa1Ka2/(Ka1Ka2+[H⁺]Ka1+[H⁺]²)
Cbβ = [B⁺] = Cb

V'を求める式
[H⁺]⁴ + {Ka1 + c'V'/(V+V')}[H⁺]³
+ {Ka1{Ka2 + (c'V'-cV)/(V+V')} - Kw}[H⁺]²
+ Ka1{Ka2(c'V'-2cV)/(V+V') - Kw}[H⁺] - Ka1Ka2Kw = 0
より
(V+V')[H⁺]⁴ + {(V+V')Ka1 + c'V'}[H⁺]³
+ {Ka1((V+V')Ka2 + (c'V'-cV)) - (V+V')Kw}[H⁺]²
+ Ka1{Ka2(c'V'-2cV) - (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 = -(-cKa1[H⁺]² - 2cKa1Ka2[H⁺] + A)
= [H⁺]cKa1([H⁺] + 2Ka2) - 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 = [H⁺]cKa1([H⁺] + 2Ka2) - A
C = c'[H⁺]{[H⁺]([H⁺] + Ka1) + Ka1Ka2} + A
V' = (B/C)V

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