pH Calculations (Divalent to Pentavalent, Mixtures)
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Perform an acid-base titration of Solution A with Solution B.
The acid valence represents the valence of free acids, such as -COOH, while the base valence represents the valence of free bases, such as -NH2.
Include acid salts, such as -ONa, in the base valence.
Include salts of bases (such as -NH₃Cl) in the acid valence.
Express the dissociation constant as the pKa of the conjugate acid.
pKa = 14 – pKb.
Select pKa values from the left in order until the sum of the acid valence and base valence is reached; ignore any remaining values.
The acid valence represents the valence of free acids, such as -COOH, while the base valence represents the valence of free bases, such as -NH2.
Include acid salts, such as -ONa, in the base valence.
Include salts of bases (such as -NH₃Cl) in the acid valence.
Express the dissociation constant as the pKa of the conjugate acid.
pKa = 14 – pKb.
Select pKa values from the left in order until the sum of the acid valence and base valence is reached; ignore any remaining values.
Let i = 1, 2, ... denote the number of a substance, and let j = 1, 2, ..., n denote the ionization number.
Let AiHai denote a compound that is “electrically neutral including its counterion.”
Let ai be the valence of the acid and bi be the valence of the base.
Kij = [AiH(ai + bi - j)^(bi - j) +] [H+] / [AiH(ai + bi - j + 1)^(bi - j + 1) +]
Ki1 = [AiH(ai + bi - 1)^(bi - 1) +] [H+] / [AiH(ai + bi)^(bi - 1) +]
[AiH(ai + bi - 1)^(bi - 1) +] = Ki1 [AiH(ai + bi)^(bi - 1) +] /[H+]
Recursively,
[AiH(ai+bi-j)^(bi-j)+] = Ki1Ki2...Kij[AiH(ai+bi)^bi+]/[H+]^j
Let the concentration before ionization be ci0. From the mass balance, we have
ci0 = [AiH(ai+bi)^bi+] ∑[j=0, ai+bi] Ki1...Kij / [H+]^j
[AiH(ai+bi)^bi+] = ci0 / Σ[j=0, ai+bi] Ki1...Kij / [H+]^j
The total charge of the AiHai ions is
ci0 × Σ[j=0, ai+bi](bi-j)(Ki1...Kij/[H+]^j) / Σ[j=0, ai+bi](Ki1...Kij/[H+]^j)
The charge balance equation is
0 = [H+] - Kw/[H+] + Σ[i=1,n]
ci0 × Σ[j=0, ai+bi](bi-j)(Ki1...Kij/[H+]^j)/Σ[j=0, ai+bi](Ki1...Kij/[H+]^j)
.
Let the volumes of each solution be V1 and V2.
Let the concentration before mixing be ci.
When present in Solution 1, ci0 = ciV1/(V1+V2)
When present in Solution 2, ci0 = ciV2/(V1+V2)
Determine the pH using the bisection method.
Let AiHai denote a compound that is “electrically neutral including its counterion.”
Let ai be the valence of the acid and bi be the valence of the base.
Kij = [AiH(ai + bi - j)^(bi - j) +] [H+] / [AiH(ai + bi - j + 1)^(bi - j + 1) +]
Ki1 = [AiH(ai + bi - 1)^(bi - 1) +] [H+] / [AiH(ai + bi)^(bi - 1) +]
[AiH(ai + bi - 1)^(bi - 1) +] = Ki1 [AiH(ai + bi)^(bi - 1) +] /[H+]
Recursively,
[AiH(ai+bi-j)^(bi-j)+] = Ki1Ki2...Kij[AiH(ai+bi)^bi+]/[H+]^j
Let the concentration before ionization be ci0. From the mass balance, we have
ci0 = [AiH(ai+bi)^bi+] ∑[j=0, ai+bi] Ki1...Kij / [H+]^j
[AiH(ai+bi)^bi+] = ci0 / Σ[j=0, ai+bi] Ki1...Kij / [H+]^j
The total charge of the AiHai ions is
ci0 × Σ[j=0, ai+bi](bi-j)(Ki1...Kij/[H+]^j) / Σ[j=0, ai+bi](Ki1...Kij/[H+]^j)
The charge balance equation is
0 = [H+] - Kw/[H+] + Σ[i=1,n]
ci0 × Σ[j=0, ai+bi](bi-j)(Ki1...Kij/[H+]^j)/Σ[j=0, ai+bi](Ki1...Kij/[H+]^j)
.
Let the volumes of each solution be V1 and V2.
Let the concentration before mixing be ci.
When present in Solution 1, ci0 = ciV1/(V1+V2)
When present in Solution 2, ci0 = ciV2/(V1+V2)
Determine the pH using the bisection method.
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Simulates acid-base titrations involving polyvalent acids, polyvalent bases, and mixtures.