12.3 Buffer Systems, Henderson-Hasselbalch Equation & Solubility Product (Ksp)

12.3 Buffer Systems, Henderson-Hasselbalch Equation & Solubility Product (Ksp)

Quick Summary: Buffer solutions resist pH alterations upon the addition of strong acids or bases via the reciprocal action of a weak acid and its conjugate base. The common ion effect suppresses weak electrolyte ionization, allowing pH to be modeled by the Henderson-Hasselbalch equation. Maximum buffer capacity occurs when weak acid and conjugate base concentrations are equimolar ($\text{pH} = \text{p}K_a$), spanning an effective buffer range of $\text{pH} = \text{p}K_a \pm 1$. In heterogeneous equilibria, the solubility product constant ($K_{sp}$) defines the dissolution limit of sparingly soluble salts. Molar solubility ($s$) is derived directly from stoichiometric ion powers. Precipitation occurs when the ion product $Q_{sp}$ exceeds $K_{sp}$, with solubility suppressed by common ions and enhanced in acidic media for salts with Brønsted-basic anions.

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1. Buffer Solutions & the Common Ion Effect

A buffer solution resists significant shifts in pH when small amounts of strong acid or base are added. An effective buffer contains comparable quantities of:

1. A weak acid ($\text{HA}$) to neutralize added hydroxide ($\text{OH}^-$).
2. Its conjugate weak base ($\text{A}^-$) to neutralize added hydronium ($\text{H}_3\text{O}^+$).

Combinations of strong acids and their salts (such as $\text{HCl} / \text{NaCl}$) cannot buffer because conjugate bases of strong acids lack proton affinity in water.

Mechanism and Common Ion Suppression

Under the common ion effect, adding a salt of the conjugate base (e.g., $\text{NaCH}_3\text{COO}$) to a weak acid ($\text{CH}_3\text{COOH}$) shifts the dissociation equilibrium to the left: $\text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CH}_3\text{COO}^-(aq)$ This suppresses ionization and maintains a reservoir of both components. Added strong acid is consumed quantitatively: $\text{A}^- + \text{H}_3\text{O}^+ \to \text{HA} + \text{H}_2\text{O}$. Added strong base is neutralized similarly: $\text{HA} + \text{OH}^- \to \text{A}^- + \text{H}_2\text{O}$. Both reactions replace strong species with weak partners, producing only minor shifts in $[\text{A}^-]/[\text{HA}]$.

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2. Henderson-Hasselbalch Equation, Buffer Capacity & Range

Rearranging $K_a = ([\text{H}_3\text{O}^+][\text{A}^-]) / [\text{HA}]$ yields the Henderson-Hasselbalch equation: $\text{pH} = \text{pK}_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = \text{pK}_a + \log_{10}\left(\frac{n_{\text{base}}}{n_{\text{acid}}}\right)$ For basic buffers: $\text{pOH} = \text{pK}_b + \log_{10}([\text{BH}^+]/[\text{B}])$.

Buffer Capacity & Effective Range

• Buffer Capacity: Measures the amount of strong acid or base a buffer can absorb before pH shifts markedly. Capacity increases with higher absolute concentrations of buffer components and is maximized when $[\text{A}^-] = [\text{HA}]$, where $\text{pH} = \text{pK}_a$.
• Buffer Range: Buffers operate effectively within the concentration ratio interval $0.10 \le [\text{A}^-]/[\text{HA}] \le 10.0$, giving a working range of $\text{pH} = \text{pK}_a \pm 1.00$.

Physiological Buffers & Preparation Guidelines

• Blood Bicarbonate Buffer: Regulates plasma at $\text{pH } 7.40 \pm 0.05$ via $\text{CO}_2(aq) + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$, with a $20:1$ $[\text{HCO}_3^-]/[\text{H}_2\text{CO}_3]$ ratio coupled to respiratory exhalation of $\text{CO}_2$.
• Intracellular Phosphate Buffer: Buffers cytosol at $\text{pH } 6.9 - 7.2$ via $\text{H}_2\text{PO}_4^- / \text{HPO}_4^{2-}$ ($\text{pK}_a = 7.20$).

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3. Worked Example: Buffer Addition Problem (Two-Step Method)

Problem: A $1.00\text{ L}$ buffer contains $0.300\text{ mol }\text{CH}_3\text{COOH}$ ($\text{pK}_a = 4.74$) and $0.300\text{ mol }\text{NaCH}_3\text{COO}$. Calculate the pH after adding $0.050\text{ mol of solid }\text{NaOH}$.

Step 1: Stoichiometry step (neutralization) Added $\text{OH}^-$ reacts completely with $\text{CH}_3\text{COOH}$: $\text{CH}_3\text{COOH} + \text{OH}^- \longrightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$

• $n(\text{CH}_3\text{COOH}) = 0.300 - 0.050 = 0.250\text{ mol}$
• $n(\text{CH}_3\text{COO}^-) = 0.300 + 0.050 = 0.350\text{ mol}$

Step 2: Equilibrium step (Henderson-Hasselbalch) $\text{pH} = 4.74 + \log_{10}\left(\frac{0.350}{0.250}\right) = 4.74 + \log_{10}(1.40) = 4.74 + 0.15 = 4.89$ The pH rises by only $0.15$ units (compared to $\text{pH } 12.70$ if added to pure water).

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4. Heterogeneous Solubility Equilibria: $K_{sp}$ & Molar Solubility

Slightly soluble salts establish heterogeneous dissolution equilibria: $\text{M}_p\text{X}_q(s) \rightleftharpoons p\,\text{M}^{m+}(aq) + q\,\text{X}^{n-}(aq) \implies K_{sp} = [\text{M}^{m+}]^p [\text{X}^{n-}]^q$ Molar solubility ($s$, in $\text{mol/L}$) relates to $K_{sp}$ based on stoichiometry:

Note: Direct comparison of $K_{sp}$ values indicates relative solubility only for salts with identical ion stoichiometries.

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5. Predicting Precipitation ($Q_{sp}$) & Solubility Factors

The ion product $Q_{sp} = [\text{M}^{m+}]_0^p [\text{X}^{n-}]_0^q$ predicts precipitation:

• $Q_{sp} < K_{sp}$: Unsaturated; no precipitate forms; more solute can dissolve.
• $Q_{sp} = K_{sp}$: Saturated solution at dynamic equilibrium.
• $Q_{sp} > K_{sp}$: Supersaturated; precipitate forms until $Q_{sp} = K_{sp}$.

Common Ion & pH Effects

• Common Ion Effect: Adding an ion present in the salt lattice shifts the equilibrium left, drastically decreasing molar solubility. For $\text{AgCl}$ ($K_{sp} = 1.8 \times 10^{-10}$), $s = 1.34 \times 10^{-5}\text{ M}$ in pure water, but drops to $1.8 \times 10^{-9}\text{ M}$ in $0.10\text{ M }\text{NaCl}$.
• pH Effect: Salts containing basic anions ($\text{OH}^-, \text{CO}_3^{2-}, \text{F}^-, \text{PO}_4^{3-}$) exhibit increased solubility in acidic solutions because hydronium protonates the anion (e.g., $\text{F}^- + \text{H}_3\text{O}^+ \rightleftharpoons \text{HF} + \text{H}_2\text{O}$), shifting the dissolution equilibrium to the right. Salts with anions of strong acids ($\text{Cl}^-, \text{Br}^-, \text{I}^-, \text{NO}_3^-$) do not dissolve more in acid.

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6. Worked Example: Precipitation Prediction ($Q_{sp}$ vs $K_{sp}$)

Problem: $100.0\text{ mL}$ of $0.020\text{ M }\text{Pb(NO}_3)_2$ is mixed with $100.0\text{ mL}$ of $0.010\text{ M }\text{NaI}$. Determine if lead(II) iodide precipitates ($K_{sp}(\text{PbI}_2) = 7.9 \times 10^{-9}$).

Step 1: Calculate diluted ion concentrations Total volume doubles to $200.0\text{ mL}$: $[\text{Pb}^{2+}]_0 = 0.020\text{ M} \times \left(\frac{100.0}{200.0}\right) = 0.010\text{ M}$ $[\text{I}^-]_0 = 0.010\text{ M} \times \left(\frac{100.0}{200.0}\right) = 0.0050\text{ M}$

Step 2: Calculate $Q_{sp}$ and compare to $K_{sp}$ $Q_{sp} = [\text{Pb}^{2+}]_0 [\text{I}^-]_0^2 = (0.010)(0.0050)^2 = (0.010)(2.5 \times 10^{-5}) = 2.5 \times 10^{-7}$ Because $Q_{sp} = 2.5 \times 10^{-7} > K_{sp} = 7.9 \times 10^{-9}$, a yellow precipitate of $\text{PbI}_2$ forms.