pH of Strong Acids/Bases

Calculating pH for strong acids and bases is the easiest type of acid-base problem. Because strong acids and bases dissociate completely, there is no equilibrium to solve - just arithmetic.

Strong Acid pH

For a monoprotic strong acid (like HCl, HBr, HI, HNO₃, HClO₃, HClO₄) at concentration c:

1. HCl → H⁺ + Cl⁻ (complete dissociation)
2. [H⁺] = c
3. pH = -log(c)

Example: What is the pH of 0.0020 M HCl?

• $[\text{H}^+] = 0.0020 \text{ M} = 2.0 \times 10^{-3}$ M
• $\text{pH} = -\log(2.0 \times 10^{-3}) = 3 - \log(2) = 3 - 0.3 =$ 2.7

Strong Base pH

For a monobasic strong base (like NaOH, KOH) at concentration c:

1. NaOH → Na⁺ + OH⁻ (complete dissociation)
2. [OH⁻] = c
3. pOH = -log(c)
4. pH = 14 - pOH

Example: What is the pH of 0.050 M NaOH?

• $[\text{OH}^-] = 0.050 \text{ M} = 5.0 \times 10^{-2}$ M
• $\text{pOH} = -\log(5.0 \times 10^{-2}) = 2 - \log(5) = 2 - 0.7 = 1.3$
• pH = 14 - 1.3 = 12.7

For dibasic strong bases (like Ba(OH)₂, Ca(OH)₂):

• [OH⁻] = 2c (two hydroxide ions per formula unit)

Example: What is the pH of 0.0010 M Ba(OH)₂?

• $[\text{OH}^-] = 2(0.0010) = 0.0020 \text{ M} = 2.0 \times 10^{-3}$ M
• pOH = 3 - log(2) = 3 - 0.3 = 2.7
• pH = 14 - 2.7 = 11.3

The Very Dilute Acid Trap

What is the pH of $10^{-8}$ M HCl?

If you naively calculate: $\text{pH} = -\log(10^{-8}) = 8$.

But wait - pH 8 is basic! A solution of acid cannot be basic, no matter how dilute.

The problem: at $10^{-8}$ M, the contribution of H⁺ from the acid is comparable to the H⁺ from water’s autoionization ($10^{-7}$ M). You must add both sources:

$[\text{H}^+]_{\text{total}} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7}$ M

$\text{pH} = -\log(1.1 \times 10^{-7}) \approx$ 6.96 (just barely acidic, as expected)

Mixing Strong Acids and Strong Bases

When a strong acid and strong base are mixed, the net reaction is neutralization:

H⁺ + OH⁻ → H₂O

After mixing, determine which species is in excess:

1. Calculate moles of H⁺ = $M_{\text{acid}}$ x $V_{\text{acid}}$
2. Calculate moles of OH⁻ = $M_{\text{base}}$ x $V_{\text{base}}$ (multiply by 2 for dibasic)
3. The excess determines pH

Example: 50.0 mL of 0.10 M HCl is mixed with 30.0 mL of 0.10 M NaOH. What is the pH?

• Moles H⁺ = (0.10)(0.050) = 0.0050 mol
• Moles OH⁻ = (0.10)(0.030) = 0.0030 mol
• Excess H⁺ = 0.0050 - 0.0030 = 0.0020 mol
• Total volume = 80.0 mL = 0.080 L
• $[\text{H}^+] = 0.0020 / 0.080 = 0.025 \text{ M} = 2.5 \times 10^{-2}$ M
• pH = 2 - log(2.5) ≈ 2 - 0.4 = 1.6