pH Change in a Buffer Solution (Weak Base + Strong Acid)
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pH Change in a Buffer Solution (Weak Base + Strong Acid)
Determine the pH by titrating a buffer solution consisting of a weak base B (initial concentration Cb) and its salt BHCl (initial concentration Ca) 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.
Both exact solutions (accurate to 10 decimal places) and approximate solutions are calculated.
The approximate solution deviates significantly from the exact solution when the basicity of B is much stronger or weaker than pKb = 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.
Both exact solutions (accurate to 10 decimal places) and approximate solutions are calculated.
The approximate solution deviates significantly from the exact solution when the basicity of B is much stronger or weaker than pKb = 7, when the buffer solution is dilute, or when the buffer capacity is exceeded.
Calculation Details
Assume that the base concentration Cb increases and the acid concentration Ca decreases in proportion to the amount of strong base added.
Assume that the acid concentration Ca increases and the base concentration Cb decreases in proportion to the amount of strong acid added.
● Approximate Solution
Ca is completely ionized.
Cb is not ionized.
Equilibrium: 10^(-pKb) = [BH+][OH-]/[B]
10^(-pKb) = Ca[OH-]/Cb
[OH-] = 10^(-pKb)(Cb/Ca)
pH = 14 − pKb + log(Cb/Ca)
● Numerical Solution
Ca is completely dissociated.
Assume that a proportion b of Cb is dissociated.
Equilibrium: 10^(-pKb) = [BH+][OH-]/[B]
10^(-pKb) = (Ca + bCb)[OH⁻] / (1 - b)Cb
Charge balance: [H⁺] + [BH⁺] = [OH⁻] + [Cl⁻]
Kw/[OH⁻] + (Ca + bCb) = [OH⁻] + Ca
Kw/[OH⁻] + bCb = [OH⁻]
bCb = [OH⁻] - Kw/[OH⁻]
(1-b)Cb = Cb - [OH⁻] + Kw/[OH⁻]
Combining these, we get
10⁻^(pKb) = (Ca + [OH⁻] - Kw/[OH⁻])[OH⁻] / (Cb - [OH⁻] + Kw/[OH⁻])
10^(-pKb) = (Ca[OH⁻] + [OH⁻]² - Kw)[OH⁻] / (Cb[OH⁻] - [OH⁻]² + Kw)
0 = (Ca[OH⁻] + [OH⁻]² - Kw)[OH⁻] - 10⁻^(pKb)(Cb[OH⁻] - [OH⁻]² + Kw)
0 = x³ + x²(Ca + 10⁻^(pKb)) + x(-Kw - 10⁻^(pKb)Cb) - 10⁻^(pKb)Kw
.
The following is the Newton method. The iteration is terminated when the change in x becomes less than 10⁻¹⁰ times the absolute value of x.
Assume that the base concentration Cb increases and the acid concentration Ca decreases in proportion to the amount of strong base added.
Assume that the acid concentration Ca increases and the base concentration Cb decreases in proportion to the amount of strong acid added.
● Approximate Solution
Ca is completely ionized.
Cb is not ionized.
Equilibrium: 10^(-pKb) = [BH+][OH-]/[B]
10^(-pKb) = Ca[OH-]/Cb
[OH-] = 10^(-pKb)(Cb/Ca)
pH = 14 − pKb + log(Cb/Ca)
● Numerical Solution
Ca is completely dissociated.
Assume that a proportion b of Cb is dissociated.
Equilibrium: 10^(-pKb) = [BH+][OH-]/[B]
10^(-pKb) = (Ca + bCb)[OH⁻] / (1 - b)Cb
Charge balance: [H⁺] + [BH⁺] = [OH⁻] + [Cl⁻]
Kw/[OH⁻] + (Ca + bCb) = [OH⁻] + Ca
Kw/[OH⁻] + bCb = [OH⁻]
bCb = [OH⁻] - Kw/[OH⁻]
(1-b)Cb = Cb - [OH⁻] + Kw/[OH⁻]
Combining these, we get
10⁻^(pKb) = (Ca + [OH⁻] - Kw/[OH⁻])[OH⁻] / (Cb - [OH⁻] + Kw/[OH⁻])
10^(-pKb) = (Ca[OH⁻] + [OH⁻]² - Kw)[OH⁻] / (Cb[OH⁻] - [OH⁻]² + Kw)
0 = (Ca[OH⁻] + [OH⁻]² - Kw)[OH⁻] - 10⁻^(pKb)(Cb[OH⁻] - [OH⁻]² + Kw)
0 = x³ + x²(Ca + 10⁻^(pKb)) + x(-Kw - 10⁻^(pKb)Cb) - 10⁻^(pKb)Kw
.
The following is the Newton method. The iteration is terminated when the change in x becomes less than 10⁻¹⁰ times the absolute value of x.
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