pH Change in a Buffer Solution (Weak Acid + Strong Base)
Author: novchemistryExecutions: 8781
※ This library was created by a member. This site is not responsible for its content.
pH Change in a Buffer Solution (Weak Acid + Strong Base)
Determine the pH by titrating a buffer solution consisting of a weak acid AH (initial concentration Ca) and a salt ANa (initial concentration Cb) of a strong base with a strong acid or strong base.
When volume changes are neglected, the concentration of the strong acid or strong base used for titration can be set to an unrealistic value, such as 100 mol/L.
We are calculating both the exact solution (accurate to 6 decimal places) and the approximate solution.
The approximate solution deviates significantly from the exact solution when the acidity of AH is much stronger or weaker than pKa = 7, when the buffer solution is dilute, or when the buffer capacity is exceeded.
When volume changes are neglected, the concentration of the strong acid or strong base used for titration can be set to an unrealistic value, such as 100 mol/L.
We are calculating both the exact solution (accurate to 6 decimal places) and the approximate solution.
The approximate solution deviates significantly from the exact solution when the acidity of AH is much stronger or weaker than pKa = 7, when the buffer solution is dilute, or when the buffer capacity is exceeded.
Calculation Details
It is assumed that the acid concentration Ca increases and the salt concentration Cb decreases in proportion to the amount of strong acid added.
It is assumed that the salt concentration Cb increases and the acid concentration Ca decreases in proportion to the amount of strong base added.
● Approximate Solution
Cb is completely dissociated.
Ca is not dissociated.
Equilibrium: 10^(-pKa) = [A-][H+]/[AH]
10^(-pKa) = Cb[H+]/Ca
[H+] = 10^(-pKa)(Ca/Cb)
pH = pKa + log(Cb/Ca)
● Numerical Solution
Cb is completely dissociated.
Assume that a proportion a of Ca is dissociated.
Equilibrium: 10^(-pKa) = [A⁻][H⁺] / [AH]
10^(-pKa) = (Cb + aCa)[H⁺] / (1 - a)Ca
Charge balance: [H+] + [Na+] = [OH-] + [A-]
[H+] + Cb = Kw/[H+] + Cb + aCa
[H+] - Kw/[H+] = aCa
Ca − [H⁺] + Kw/[H⁺] = (1 − a)Ca
Combining these, we get
10^(−pKa) = (Cb + [H⁺] − Kw/[H⁺])[H⁺] / (Ca − [H⁺] + Kw/[H⁺])
10^(-pKa) = (Cb[H+] + [H+]₂ - Kw)[H+] / (Ca[H+] - [H+]₂ + Kw)
0 = (Cb[H+] + [H+]² - Kw)[H+] - 10⁻^(pKa)(Ca[H+] - [H+]² + Kw)
0 = x³ + x²(Cb + 10⁻^(pKa)) + x(-Kw - 10⁻^(pKa)Ca) - 10⁻^(pKa)Kw
.
The following is the Newton method. The calculation is terminated when the change in x becomes less than 10^(-10) times the absolute value of x.
Regarding when the buffer is disrupted:
C = Ca + Cb
Let C' be the concentration of excess strong acid.
Charge balance: [H+] = [OH-] + [A-] + [Cl-]
[H+] = Kw/[H+] + aC + C'
a = ([H+] - Kw/[H+] - C')/C
10^(-pKa) = [A-][H+] / [AH]
10^(-pKa) = a[H+] / (1 - a)
10^(-pKa)(1 - a) = a[H+]
10^(-pKa) = a([H+] + 10^(-pKa))
a = 10^(-pKa) / ([H+] + 10^(-pKa)) = ([H+] - Kw/[H+] - C') / C
10^(-pKa)C = ([H+] - Kw / [H+] - C')([H+] + 10^(-pKa))
10^(-pKa)C[H+] = ([H+]² - Kw - C'[H+])([H+] + 10^(-pKa))
0 = x³ + x²(10^(-pKa) - C') + x(-10^(-pKa)C - 10^(-pKa)C' - Kw) - 10^(-pKa)Kw
Since we can simply substitute
C + C' = Ca
and
-C' = Cb,
no special handling is required.
It is assumed that the acid concentration Ca increases and the salt concentration Cb decreases in proportion to the amount of strong acid added.
It is assumed that the salt concentration Cb increases and the acid concentration Ca decreases in proportion to the amount of strong base added.
● Approximate Solution
Cb is completely dissociated.
Ca is not dissociated.
Equilibrium: 10^(-pKa) = [A-][H+]/[AH]
10^(-pKa) = Cb[H+]/Ca
[H+] = 10^(-pKa)(Ca/Cb)
pH = pKa + log(Cb/Ca)
● Numerical Solution
Cb is completely dissociated.
Assume that a proportion a of Ca is dissociated.
Equilibrium: 10^(-pKa) = [A⁻][H⁺] / [AH]
10^(-pKa) = (Cb + aCa)[H⁺] / (1 - a)Ca
Charge balance: [H+] + [Na+] = [OH-] + [A-]
[H+] + Cb = Kw/[H+] + Cb + aCa
[H+] - Kw/[H+] = aCa
Ca − [H⁺] + Kw/[H⁺] = (1 − a)Ca
Combining these, we get
10^(−pKa) = (Cb + [H⁺] − Kw/[H⁺])[H⁺] / (Ca − [H⁺] + Kw/[H⁺])
10^(-pKa) = (Cb[H+] + [H+]₂ - Kw)[H+] / (Ca[H+] - [H+]₂ + Kw)
0 = (Cb[H+] + [H+]² - Kw)[H+] - 10⁻^(pKa)(Ca[H+] - [H+]² + Kw)
0 = x³ + x²(Cb + 10⁻^(pKa)) + x(-Kw - 10⁻^(pKa)Ca) - 10⁻^(pKa)Kw
.
The following is the Newton method. The calculation is terminated when the change in x becomes less than 10^(-10) times the absolute value of x.
Regarding when the buffer is disrupted:
C = Ca + Cb
Let C' be the concentration of excess strong acid.
Charge balance: [H+] = [OH-] + [A-] + [Cl-]
[H+] = Kw/[H+] + aC + C'
a = ([H+] - Kw/[H+] - C')/C
10^(-pKa) = [A-][H+] / [AH]
10^(-pKa) = a[H+] / (1 - a)
10^(-pKa)(1 - a) = a[H+]
10^(-pKa) = a([H+] + 10^(-pKa))
a = 10^(-pKa) / ([H+] + 10^(-pKa)) = ([H+] - Kw/[H+] - C') / C
10^(-pKa)C = ([H+] - Kw / [H+] - C')([H+] + 10^(-pKa))
10^(-pKa)C[H+] = ([H+]² - Kw - C'[H+])([H+] + 10^(-pKa))
0 = x³ + x²(10^(-pKa) - C') + x(-10^(-pKa)C - 10^(-pKa)C' - Kw) - 10^(-pKa)Kw
Since we can simply substitute
C + C' = Ca
and
-C' = Cb,
no special handling is required.
No customer feedback for this calculation yet.