Master pH calculations and understand acid-base equilibrium through interactive lessons. Learn to calculate pH, pOH, and hydrogen ion concentrations with real-world applications.
Interactive Learning: Use our pH Calculator alongside this tutorial for hands-on practice.
pH stands for "potential of Hydrogen" and measures the acidity or alkalinity of a solution. The pH scale ranges from 0 to 14, with 7 being neutral.
pH 0-6.9
More H⁺ ions
pH 7.0
Equal H⁺ and OH⁻
pH 7.1-14
More OH⁻ ions
The most fundamental pH calculation involves converting hydrogen ion concentration to pH using the logarithmic relationship.
Identify [H⁺] concentration
Usually given in scientific notation (e.g., 1.0 × 10⁻³ M)
Apply the pH formula
pH = -log[H⁺]
Calculate using logarithms
Use a calculator or logarithm tables
Given: [H⁺] = 1.0 × 10⁻⁴ M
Step 1: pH = -log(1.0 × 10⁻⁴)
Step 2: pH = -log(10⁻⁴)
Step 3: pH = -(-4) = 4
Answer: pH = 4 (acidic solution)
Given: [H⁺] = 3.5 × 10⁻⁶ M
Step 1: pH = -log(3.5 × 10⁻⁶)
Step 2: pH = -[log(3.5) + log(10⁻⁶)]
Step 3: pH = -[0.544 + (-6)] = -(-5.456)
Answer: pH = 5.46
Water undergoes autoionization, creating both H⁺ and OH⁻ ions. This relationship is fundamental to understanding pH and pOH.
H₂O ⇌ H⁺ + OH⁻
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ (at 25°C)
pOH = -log[OH⁻]
pH + pOH = 14 (at 25°C)
[H⁺] × [OH⁻] = 1.0 × 10⁻¹⁴
Given: [OH⁻] = 2.0 × 10⁻³ M
Step 1: pOH = -log(2.0 × 10⁻³) = 2.70
Step 2: pH = 14 - pOH = 14 - 2.70 = 11.30
Answer: pH = 11.30 (basic solution)
Strong acids and bases completely ionize in water, making pH calculations straightforward. The concentration of the acid or base directly determines the ion concentrations.
*Diprotic acid
*Diprotic base
Calculate pH of 0.01 M HCl solution
HCl → H⁺ + Cl⁻ (complete ionization)
[H⁺] = 0.01 M = 1.0 × 10⁻² M
pH = -log(1.0 × 10⁻²) = 2
Answer: pH = 2
Calculate pH of 0.005 M NaOH solution
NaOH → Na⁺ + OH⁻ (complete ionization)
[OH⁻] = 0.005 M = 5.0 × 10⁻³ M
pOH = -log(5.0 × 10⁻³) = 2.30
pH = 14 - 2.30 = 11.70
Answer: pH = 11.70
Weak acids only partially ionize in water, requiring equilibrium calculations to determine pH. The acid dissociation constant (Ka) quantifies the strength of weak acids.
HA ⇌ H⁺ + A⁻
Ka = [H⁺][A⁻] / [HA]
Acetic acid (CH₃COOH): Ka = 1.8 × 10⁻⁵
Formic acid (HCOOH): Ka = 1.8 × 10⁻⁴
Hydrofluoric acid (HF): Ka = 6.8 × 10⁻⁴
Carbonic acid (H₂CO₃): Ka₁ = 4.3 × 10⁻⁷
For calculating weak acid pH:
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| Initial (I) | C | 0 | 0 |
| Change (C) | -x | +x | +x |
| Equilibrium (E) | C-x | x | x |
Buffer solutions resist changes in pH when small amounts of acid or base are added. They consist of a weak acid and its conjugate base (or weak base and its conjugate acid).
pH = pKa + log([A⁻]/[HA])
where pKa = -log(Ka)
Buffer: 0.1 M CH₃COOH + 0.1 M CH₃COONa
Ka for acetic acid = 1.8 × 10⁻⁵
pKa = -log(1.8 × 10⁻⁵) = 4.74
pH = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74
Buffer pH = 4.74
Calculate the pH of 0.025 M HNO₃ solution.
Remember: HNO₃ is a strong acid that completely ionizes.
What is the pH of 0.008 M Ba(OH)₂ solution?
Hint: Ba(OH)₂ produces 2 OH⁻ ions per formula unit.
A solution has pH = 3.5. What is [H⁺]?
Use the relationship: [H⁺] = 10⁻ᵖᴴ