Mastering Acid-Base (pH) and Solubility (Ksp) Equilibrium Calculations
Chemical equilibrium is often the defining hurdle in general chemistry. It represents the shift from viewing reactions as simple "start-to-finish" processes to understanding them as dynamic systems where forward and reverse reaction rates balance out.
Whether you are calculating the acidity of a weak preservative or determining how much toxic metal dissolves in a water supply, the underlying math remains the same: the Law of Mass Action. In this guide, we will break down the three most critical applications of equilibrium: Acid-Base ($K_a/K_b$), Salt Hydrolysis, and Solubility ($K_{sp}$).
The Ion-Product of Water ($K_w$) and $\text{pH} / \text{pOH}$
Before diving into complex molecules, we must understand the solvent: water. Water is not chemically inert; it undergoes autoionization, where water molecules transfer protons to one another.
$$ 2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq) $$The equilibrium constant for this process is called the ion-product constant of water ($K_w$). At $25^\circ\text{C}$:
$$ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} $$This relationship holds true for any aqueous solution. If you know the concentration of hydronium ions $[H_3O^+]$, you automatically know the concentration of hydroxide ions $[OH^-]$. This gives rise to the pH and pOH scales:
- pH $= -\log[H_3O^+]$
- pOH $= -\log[OH^-]$
- Relationship: $\text{pH} + \text{pOH} = 14.00$ (at $25^\circ\text{C}$)
Weak Acid/Base Equilibria ($K_a$ and $K_b$)
Strong acids (like HCl) dissociate completely. Weak acids (like acetic acid or HF) do not. To find the pH of a weak acid, we must quantify how "weak" it is using the Acid Dissociation Constant ($K_a$).
The general equilibrium for a weak acid (HA) is:
$$ HA(aq) \rightleftharpoons H^+(aq) + A^-(aq) $$ $$ K_a = \frac{[H^+][A^-]}{[HA]} $$Technique: The ICE Table and the Small $x$ Approximation
To solve these problems systematically, we use an ICE table (Initial, Change, Equilibrium). Let's look at a standard case using Hydrocyanic Acid (HCN).
Illustrative Example: pH of a Weak Acid
Problem: Calculate the pH of a $0.247 \text{ M}$ solution of hydrocyanic acid (HCN). Given $K_a = 6.2 \times 10^{-10}$.
Step 1: Set up the ICE Table.
We assume the dissociation produces $x$ amount of $H^+$.
| Species | [HCN] | [H+] | [CN-] |
|---|---|---|---|
| Initial | 0.247 | 0 | 0 |
| Change | $-x$ | $+x$ | $+x$ |
| Equilibrium | $0.247 - x$ | $x$ | $x$ |
Step 2: Write the Equilibrium Expression.
$$ 6.2 \times 10^{-10} = \frac{(x)(x)}{0.247 - x} $$Step 3: Apply the Approximation.
Because $K_a$ is extremely small ($10^{-10}$), the amount dissociated ($x$) will be tiny compared to the initial concentration ($0.247$). We can assume $0.247 - x \approx 0.247$.
Step 4: Calculate pH.
$$ \text{pH} = -\log(1.24 \times 10^{-5}) \approx \mathbf{4.91} $$When the Approximation Fails (Quadratic Method)
If the acid is stronger (larger $K_a$) or the solution is very dilute, the "$x$ is small" assumption fails (usually if $x > 5\%$ of the initial concentration). In these cases, you must solve the full quadratic equation.
Case Study: Nitrous Acid ($HNO_2$)
Consider a $0.010 \text{ M}$ solution of $HNO_2$ ($K_a = 4.6 \times 10^{-4}$).
- Setup: $K_a = \frac{x^2}{0.010 - x} = 4.6 \times 10^{-4}$
- Rearrange: $x^2 + (4.6 \times 10^{-4})x - (4.6 \times 10^{-6}) = 0$
- Quadratic Formula: Using $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$, we find the positive root is $x = 1.93 \times 10^{-3} \text{ M}$.
- Result: $\text{pH} = -\log(1.93 \times 10^{-3}) = \mathbf{2.71}$.
Note: If we had used the approximation here, we would have calculated a pH of 2.67, introducing a significant error.
Buffer Solutions and Salt Hydrolysis
Acids and bases don't always come in pure forms. Often, we deal with mixtures (buffers) or salts that react with water (hydrolysis).
Salt Hydrolysis: The Amphiprotic Case
Salts formed from weak acids and weak bases can be tricky. Some ions are amphiprotic, meaning they can act as either an acid or a base. A classic example is Sodium Bicarbonate ($NaHCO_3$), commonly known as baking soda.
The bicarbonate ion ($HCO_3^-$) has two choices:
- Act as an acid: $HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}$ (governed by $K_{a2}$)
- Act as a base: $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$ (governed by $K_b$)
To find the pH of an amphiprotic salt solution, we generally don't need an ICE table. Instead, the pH is determined by the average of the $pK_a$ values surrounding the species.
$$ \text{pH} \approx \frac{pK_{a1} + pK_{a2}}{2} $$Practice Problem: pH of Baking Soda ($NaHCO_3$)
Find the pH of a $0.488 \text{ M}$ solution of $NaHCO_3$. Carbonic acid values are $K_{a1} = 4.2 \times 10^{-7}$ and $K_{a2} = 4.8 \times 10^{-11}$.
Solution:
- First, calculate the pK values:
- $pK_{a1} = -\log(4.2 \times 10^{-7}) = 6.377$
- $pK_{a2} = -\log(4.8 \times 10^{-11}) = 10.319$
- Apply the amphiprotic formula: $$ \text{pH} = \frac{6.377 + 10.319}{2} = \mathbf{8.35} $$
Note how the concentration ($0.488 \text{ M}$) essentially cancels out in this approximation, provided the concentration is sufficiently high.
Solubility Equilibria and the $K_{sp}$ Constant
Solubility is effectively a battle between the lattice energy of a solid and the hydration energy of the ions. The equilibrium constant for a solid dissolving is the Solubility Product Constant ($K_{sp}$).
For a salt $M_x A_y(s)$:
$$ M_x A_y(s) \rightleftharpoons xM^{y+}(aq) + yA^{x-}(aq) $$ $$ K_{sp} = [M^{y+}]^x [A^{x-}]^y $$From Solubility to $K_{sp}$
Molar solubility ($s$) is the maximum number of moles of solute that can dissolve in 1 Liter of solvent. Students often confuse $s$ with $K_{sp}$, but they are distinct: $s$ is a concentration (M), while $K_{sp}$ is the equilibrium constant.
Example: Lanthanum (III) Fluoride ($LaF_3$)
Given a solubility of $1.0 \times 10^{-5} \text{ g/100 mL}$ and molar mass $195.9 \text{ g/mol}$, find $K_{sp}$.
- Convert Solubility to Molarity ($s$): $$ 1.0 \times 10^{-5} \frac{\text{g}}{100\text{mL}} \times \frac{1000\text{mL}}{1\text{L}} = 1.0 \times 10^{-4} \text{ g/L} $$ $$ s = \frac{1.0 \times 10^{-4} \text{ g/L}}{195.9 \text{ g/mol}} \approx 5.10 \times 10^{-7} \text{ M} $$
- Define Ion Concentrations:
Equation: $LaF_3(s) \rightleftharpoons La^{3+} + 3F^-$
$[La^{3+}] = s$
$[F^-] = 3s$ - Calculate $K_{sp}$: $$ K_{sp} = [s][3s]^3 = 27s^4 $$ $$ K_{sp} = 27(5.10 \times 10^{-7})^4 \approx \mathbf{1.8 \times 10^{-24}} $$
Linking Solubility and pH
Many slightly soluble salts contain basic anions (like $OH^-$, $CO_3^{2-}$, or $F^-$). For metal hydroxides, the solubility directly determines the pH of the saturated solution.
Example: Saturated Iron(II) Hydroxide $Fe(OH)_2$
Given $K_{sp} = 4.87 \times 10^{-17}$, what is the pH?
- Step 1: $K_{sp} = [Fe^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$.
- Step 2: Solve for molar solubility $s$. $$ s = \sqrt[3]{\frac{4.87 \times 10^{-17}}{4}} \approx 2.30 \times 10^{-6} \text{ M} $$
- Step 3: Find $[OH^-]$. $$ [OH^-] = 2s = 4.60 \times 10^{-6} \text{ M} $$
- Step 4: Convert to pH. $$ \text{pOH} = -\log(4.60 \times 10^{-6}) = 5.34 $$ $$ \text{pH} = 14.00 - 5.34 = \mathbf{8.66} $$
Key Takeaways for Equilibrium Calculations
- The Process is Universal: Whether it's $K_a$, $K_b$, or $K_{sp}$, the logic is always Products over Reactants (raised to their coefficients).
- Check Your Assumptions: The approximation $C_{initial} - x \approx C_{initial}$ saves time, but it only works when $K$ is small. If $x$ is $>5\%$ of the initial concentration, use the quadratic formula.
- Watch Your Units: Always convert solubility from grams/Liter to Molarity (mol/L) before plugging values into a $K_{sp}$ expression.
- Temperature Matters: Equilibrium constants change with temperature. Unless stated otherwise, assume standard $25^\circ\text{C}$ conditions ($pH + pOH = 14$).