3.3 pH scale and calculations for strong and weak acids/bases

pH and the Concentration of Hydronium Ions

Definition and significance of pH

The pH scale is a logarithmic scale that measures how acidic or basic a solution is. Because hydrogen ion concentrations span many orders of magnitude, the log scale compresses those numbers into a manageable 0–14 range.

• pH is defined as the negative base-10 logarithm of the hydronium ion concentration: $pH = -\log_{10}[H_3O^+]$
• The scale runs from 0 to 14 at 25°C, with 7 being neutral (pure water).
  • Lower pH = more acidic (lemon juice ≈ 2, vinegar ≈ 3)
  • Higher pH = more basic (baking soda ≈ 8, bleach ≈ 13)

Because the scale is logarithmic, each whole-number change in pH represents a tenfold change in $[H_3O^+]$. A solution at pH 3 has ten times more hydronium ions than one at pH 4.

The autoionization of water

In pure water at 25°C, water molecules donate and accept protons from each other in small amounts:

$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}$

This produces equal concentrations of hydronium and hydroxide ions:

$[H_3O^+] = [OH^-] = 1.0 \times 10^{-7} \text{ M}$

That gives a neutral pH of 7. From here, the classification is straightforward:

• Acidic solutions: $[H_3O^+] > 1.0 \times 10^{-7}$ M, so $pH < 7$
• Basic solutions: $[H_3O^+] < 1.0 \times 10^{-7}$ M, so $pH > 7$

Calculating pH for Strong and Weak Acids and Bases

pH calculations for strong acids and bases

Strong acids and bases completely dissociate in water. That means every molecule that dissolves breaks apart into ions, so the math is direct.

Strong acids (e.g., HCl, $HNO_3$, $H_2SO_4$ first proton):

$HA_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + A^-_{(aq)}$

Since dissociation is 100%, $[H_3O^+]$ equals the initial acid concentration.

Strong bases (e.g., NaOH, KOH):

$MOH_{(aq)} \rightarrow M^+_{(aq)} + OH^-_{(aq)}$

Here $[OH^-]$ equals the initial base concentration. For bases like $Ba(OH)_2$ that release two $OH^-$ per formula unit, $[OH^-] = 2 \times$ the initial concentration.

Steps for a strong acid:

1. Identify the initial concentration of the acid (e.g., 0.010 M HCl).
2. Set $[H_3O^+]$ equal to that concentration: $[H_3O^+] = 0.010$ M.
3. Calculate pH: $pH = -\log_{10}(0.010) = 2.00$.

Steps for a strong base:

1. Identify the initial concentration (e.g., 0.0050 M NaOH).
2. Set $[OH^-] = 0.0050$ M.
3. Calculate pOH: $pOH = -\log_{10}(0.0050) = 2.30$.
4. Convert to pH: $pH = 14 - 2.30 = 11.70$.

pH determination for weak acids and bases

Weak acids and bases only partially dissociate, so you can't just set the ion concentration equal to the initial concentration. Instead, you need the equilibrium constant and an ICE table.

Weak acid equilibrium (e.g., acetic acid $CH_3COOH$, $K_a = 1.8 \times 10^{-5}$):

$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}$

$K_a = \frac{[H_3O^+][A^-]}{[HA]}$

Steps for a weak acid (using the approximation method):

1. Write the $K_a$ expression and set up an ICE table. Let $x = [H_3O^+] = [A^-]$ at equilibrium.

2. The equilibrium concentration of HA is $(C_0 - x)$, where $C_0$ is the initial concentration.

3. Substitute into the $K_a$ expression: $K_a = \frac{x^2}{C_0 - x}$

4. Apply the 5% approximation: if $C_0$ is much larger than $K_a$ (roughly $C_0 / K_a > 400$), assume $C_0 - x \approx C_0$. This simplifies the equation to: $x = \sqrt{K_a \cdot C_0}$

5. Check the approximation: verify that $\frac{x}{C_0} \times 100\% < 5\%$. If it's not, go back and solve the full quadratic.

6. Calculate pH: $pH = -\log_{10}(x)$

Example: Find the pH of 0.10 M acetic acid ($K_a = 1.8 \times 10^{-5}$).

• $x = \sqrt{(1.8 \times 10^{-5})(0.10)} = \sqrt{1.8 \times 10^{-6}} = 1.3 \times 10^{-3}$ M
• Check: $(1.3 \times 10^{-3} / 0.10) \times 100\% = 1.3\%$ ✓ (under 5%)
• $pH = -\log_{10}(1.3 \times 10^{-3}) = 2.89$

Weak base equilibrium (e.g., ammonia $NH_3$, $K_b = 1.8 \times 10^{-5}$):

$B_{(aq)} + H_2O_{(l)} \rightleftharpoons BH^+_{(aq)} + OH^-_{(aq)}$

$K_b = \frac{[BH^+][OH^-]}{[B]}$

The process mirrors the weak acid calculation, but you solve for $[OH^-]$ first:

1. Let $x = [OH^-] = [BH^+]$ and set up the ICE table.
2. Apply the same approximation: $x = \sqrt{K_b \cdot C_0}$
3. Check the 5% rule.
4. Calculate pOH: $pOH = -\log_{10}(x)$
5. Convert: $pH = 14 - pOH$

Converting Between pH, pOH, [H₃O⁺], and [OH⁻]

These four quantities are all connected. If you know any one of them, you can find the other three. Here are the key relationships:

The pH–pOH relationship (at 25°C):

$pH + pOH = 14$

• Given pH, find pOH: $pOH = 14 - pH$
• Given pOH, find pH: $pH = 14 - pOH$

The ion product of water:

$K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \text{ at 25°C}$

• Given $[H_3O^+]$, find $[OH^-]$: $[OH^-] = \frac{K_w}{[H_3O^+]}$
• Given $[OH^-]$, find $[H_3O^+]$: $[H_3O^+] = \frac{K_w}{[OH^-]}$

Converting between pH and concentration:

• $pH = -\log_{10}[H_3O^+]$ and $[H_3O^+] = 10^{-pH}$
• $pOH = -\log_{10}[OH^-]$ and $[OH^-] = 10^{-pOH}$

Quick example: You're given pH = 4.50. Find everything else.

1. $[H_3O^+] = 10^{-4.50} = 3.2 \times 10^{-5}$ M

2. $pOH = 14 - 4.50 = 9.50$

3. $[OH^-] = 10^{-9.50} = 3.2 \times 10^{-10}$ M

You can also verify: $(3.2 \times 10^{-5})(3.2 \times 10^{-10}) = 1.0 \times 10^{-14}$ ✓