中和滴定曲線 (2回目)

中和滴定曲線 (2回目)

2026/1/6(火)

中和滴定曲線 (2回目)

(neutralization titration)

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

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 = 1
[A^n-] = CaKa/Σ_j=0ⁿ([H⁺]^jΠ_i=1^n-jKa_i)より
[A^-] = CaKa/Σ_j=0¹([H⁺]^jΠ_i=1^1-jKa_i)
= CaKa/([H⁺]⁰Π_i=1^1-0Ka_i+[H⁺]¹Π_i=1^1-1Ka_i)
= CaKa/(Ka₁+[H⁺]) = CaKa/(Ka+[H⁺])

Caα = [A^-] = CaKa/(Ka+[H⁺])
x = Σ(k[H_n-kA^k-]) = [A^-]

Cbβ = [B⁺] = CbKb/(Kb+[OH^-]) = [H⁺]CbKb/([H⁺]Kb+Kw)
y = Σ(k[B(OH)_m-k^k+]) = [B⁺]

-(x-y)[H⁺] = -([A^-]-[B⁺])[H⁺]
= [H⁺]²CbKb/([H⁺]Kb+Kw) - [H⁺]CaKa/(Ka+[H⁺])
= {[H⁺]²CbKb(Ka+[H⁺])-[H⁺]CaKa([H⁺]Kb+Kw)}/{([H⁺]Kb+Kw)(Ka+[H⁺])}
= ([H⁺]²CbKaKb+[H⁺]³CbKb-[H⁺]²CaKaKb-[H⁺]CaKaKw)
/ ([H⁺]KaKb+KaKw+[H⁺]²Kb+[H⁺]Kw)
= {CbKb[H⁺]³+KaKb(Cb-Ca)[H⁺]²-CaKaKw[H⁺]}
/ (Kb[H⁺]²+(KaKb+Kw)[H⁺]+KaKw)
を
[H⁺]² - (x - y)[H⁺] - Kw = 0
に代入
(Kb[H⁺]²+(KaKb+Kw)[H⁺]+KaKw)[H⁺]²
+ {CbKb[H⁺]³+KaKb(Cb-Ca)[H⁺]²-CaKaKw[H⁺]}
- (Kb[H⁺]²+(KaKb+Kw)[H⁺]+KaKw)Kw
= Kb[H⁺]⁴+(KaKb+Kw)[H⁺]³+KaKw[H⁺]²
+ CbKb[H⁺]³+KaKb(Cb-Ca)[H⁺]²-CaKaKw[H⁺]
- KbKw[H⁺]²-Kw(KaKb+Kw)[H⁺]-KaKw²
= Kb[H⁺]⁴+{Kw+(Ka+Cb)Kb}[H⁺]³+{(Ka-Kb)Kw+(Cb-Ca)KaKb}[H⁺]²
-{Kw+(Kb+Ca)Ka+Kw}Kw[H⁺]-KaKw² = 0

Kb[H⁺]⁴+{Kw+(Ka+Cb)Kb}[H⁺]³+{(Ka-Kb)Kw+(Cb-Ca)KaKb}[H⁺]²
-{Kw+(Kb+Ca)Ka}Kw[H⁺]-KaKw² = 0

Ca = cV/(V+V') , Cb = c'V'/(V+V')
Caα = [A^-] = CaKa/(Ka+[H⁺])
Cbβ = [B⁺] = [H⁺]CbKb/([H⁺]Kb+Kw)

V'を求める式
Kb[H⁺]⁴+{Kw+(Ka+c'B'/(V+V'))Kb}[H⁺]³
+ {(Ka-Kb)Kw+(c'V'-cV)/(V+V')KaKb}[H⁺]²
-{Kw+(Kb+cV/(V+V'))Ka}Kw[H⁺]-KaKw² = 0
より
(V+V')Kb[H⁺]⁴+{(V+V')Kw+(V+V')KaKb+c'V'Kb}[H⁺]³
+{(V+V')(Ka-Kb)Kw+(c'V'-cV)KaKb}[H⁺]²
-{(V+V')Kw+(V+V')KaKb+cVKa}Kw[H⁺]-KaKw² = 0

A = Kb[H⁺]⁴+Kw[H⁺]³+KaKb[H⁺]³+(Ka-Kb)Kw[H⁺]²-Kw²[H⁺]-KaKbKw[H⁺]
= Kb[H⁺]⁴+(Kw+KaKb)[H⁺]³+(Ka-Kb)Kw[H⁺]²-(Kw+KaKb)Kw[H⁺]
= [H⁺]{Kb[H⁺]³+(Kw+KaKb)[H⁺]²+(Ka-Kb)Kw[H⁺]-(Kw+KaKb)Kw}
= [H⁺]{[H⁺]{[H⁺]{Kb[H⁺]+(Kw+KaKb)}+(Ka-Kb)Kw}-(Kw+KaKb)Kw}
B = cKaKb[H⁺]²+cKaKw[H⁺]+KaKw² - A = (Kb[H⁺]+Kw)cKa[H⁺]+KaKw² - A
C = c'Kb[H⁺]³+c'KaKb[H⁺]² + A = ([H⁺]+Ka)c'Kb[H⁺]² + A

A = [H⁺]{[H⁺]{[H⁺]{Kb[H⁺]+(Kw+KaKb)}+(Ka-Kb)Kw}-(Kw+KaKb)Kw}
B = (Kb[H⁺]+Kw)cKa[H⁺]+KaKw² - A
C = ([H⁺]+Ka)c'Kb[H⁺]² + A
V' = (B/C)V

以下、Excelによる描画