Theory — Acids, Bases, Buffers and pH

Strong vs Weak; Electrolytes

A strong acid or base dissociates completely in water, so it is a strong electrolyte (conducts well). A weak acid or base only partially dissociates, making it a weak electrolyte. The degree of dissociation controls both the pH and the conductivity.

pH and pOH pH = −log₁₀[H⁺]  ·  pOH = −log₁₀[OH⁻]
[H⁺] = 10^(−pH)  ·  [OH⁻] = 10^(−pOH)
pH + pOH = 14  (at 25 °C)
Strong acid: [H⁺] = concentration · Strong base: [OH⁻] = concentration

Weak Acids and Bases — Ka and Kb

Weak acids and bases reach an equilibrium. The acid dissociation constant Ka (or base constant Kb) measures the extent of dissociation. For a weak acid HA at concentration C, solving Ka = x²/(C − x) gives [H⁺] = x.

Weak Acid / Base Equilibrium HA ⇌ H⁺ + A⁻  Ka = [H⁺][A⁻] / [HA]
B + H₂O ⇌ BH⁺ + OH⁻  Kb = [BH⁺][OH⁻] / [B]
pKa = −log Ka  ·  pKb = −log Kb  ·  pKa + pKb = 14
% dissociation = ([H⁺] or [OH⁻]) ÷ C × 100

Buffers — Henderson–Hasselbalch

A buffer resists changes in pH and is made from a weak acid and its conjugate base (or a weak base and its conjugate acid).

Buffer pH pH = pKa + log₁₀( [A⁻] / [HA] )
When [A⁻] = [HA], pH = pKa

Strong Electrolyte

Full dissociation; many free ions; conducts strongly; pH from full concentration.

Weak Electrolyte

Partial dissociation; few free ions; conducts weakly; pH from Ka or Kb.

Buffer

Weak acid + conjugate base; resists pH change; pH set by the ratio.

QuantityRelationship
pH−log[H⁺]
pOH−log[OH⁻]
[H⁺]10^(−pH)
Weak acid [H⁺]√(Ka·C) (approx) or quadratic
BufferpKa + log([A⁻]/[HA])

Apparatus

The equipment a real acid-base titration experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

delivers titrant
Burette
Delivers titrant drop by drop to find the equivalence point.
7.00measures pH
pH meter
Tracks the pH continuously to plot the titration curve.
shows pH by colour
Indicator
Changes colour near the endpoint of the neutralization.
reaction flask
Erlenmeyer flask
Holds the analyte solution being titrated.
exact volume
Volumetric flask
Prepares the buffer and standard solutions to an exact volume.
holds solutions
Beaker
Holds and mixes the acid, base, and buffer solutions.

Instructions — Running the Virtual Experiment

The simulation has four sections. Complete each one and record your results, with screenshots, in your lab report.

Part 1 — Solutions (Solutions tab)
1
Open Simulation → Solutions. Select a solution type and concentration. Before running it, calculate the expected pH yourself (for a strong acid pH = −log[H⁺]; for a strong base pH = 14 − pOH). Then read the measured pH from the simulation and record both your calculated value and the lab value.
2
Do this for the full set of tests (strong and weak acids and bases at 0.001, 0.01, 0.1, and 1 mol/L). For each, also note the dissociation and the electrolyte strength. For the strong acid and strong base, compare your calculated pH with the measured pH.
Part 2 — pH Calculator (pH ↔ Concentration tab)
1
Open pH ↔ Concentration. Use the drop-down to choose what you have (pH, [H⁺], or [OH⁻]); only that box is active and the others are greyed out. Work out the conversion by hand first, then enter your value and click Convert to check.
2
Do two conversions (for example, one starting from pH and one starting from a concentration). Record your own calculation and the value the lab gives for each.
Part 3 — Ka / Kb (Weak Acid & Base tab)
1
Open Ka / Kb. Choose a weak acid or base and set Ka/Kb and the concentration. Calculate the pH and percent dissociation yourself first (solve Ka = x²/(C − x) for x), then click Calculate and record both your answer and the lab's answer.
2
Do two examples (for instance a weak acid and a weak base, or two different concentrations).
Part 4 — Buffers (Buffer tab)
1
Open Buffer. Set the pKa and the concentrations of the weak acid and its conjugate base. Calculate the buffer pH yourself first using pH = pKa + log([A⁻]/[HA]), then click Calculate and record both values.
2
Do two examples with different ratios (for example equal concentrations, and then more conjugate base than acid) and note how the pH changes.

Simulation — Acids, Bases, Buffers and pH

Acid–Base Virtual LabSolutions · pH · Ka/Kb · Buffers
0247101214
Strong electrolyte

Choose a solution

Measurement
SolutionStrong acid
[H⁺]1.0e-1 M
[OH⁻]1.0e-13 M
Measured pH1.00
Dissociation100%
Strong acids fully dissociate.
0247101214

Enter a value in any one field and click its Convert button. The others are calculated from pH + pOH = 14 and [H⁺][OH⁻] = 10⁻¹⁴.

Convert between pH and concentration

Result
pH—
pOH—
[H⁺]—
[OH⁻]—
Acidic if pH < 7, basic if pH > 7.
0247101214

For a weak acid or base of concentration C, the program solves the equilibrium exactly: x²/(C − x) = K, with [H⁺] (or [OH⁻]) = x.

Weak acid or base

Result
[H⁺]—
pKa—
% dissociation—
pH—
Choose acid or base, set K and C, then calculate.
0247101214

A buffer of a weak acid (HA) and its conjugate base (A⁻). pH = pKa + log([A⁻]/[HA]). When the two concentrations are equal, pH = pKa.

Buffer composition

Result
ratio [A⁻]/[HA]1.00
log(ratio)0.00
Buffer pH4.74
When [A⁻] = [HA], pH = pKa.

Team Questions

Question 1. What is the pH of a 0.01 mol/L strong acid (full dissociation)? (Type to 2 decimals)
Question 2. What is the pH of a 0.1 mol/L strong base? (pOH = −log[OH⁻], pH = 14 − pOH; type to 2 decimals)
Question 3. If [H⁺] = 1.0 × 10⁻⁴ mol/L, what is the pH? (Type to 2 decimals)
Question 4. A solution has pH = 5.0. What is [H⁺] in mol/L? (Type, e.g. 1e-5)
Question 5. For a weak acid with Ka = 1.8 × 10⁻⁵, what is the pKa? (Type to 2 decimals)
Question 6. A buffer has pKa = 4.74 with equal concentrations of acid and conjugate base. What is the pH? (Type to 2 decimals)
Question 7 — Challenge. Why does a 0.1 mol/L weak acid have a higher pH than a 0.1 mol/L strong acid? (Answer in a phrase)

Example Lab Report

Sample report demonstrating the expected format. Include your full data table of all sixteen solution tests, the theoretical comparison for the strong acid and base, and labelled screenshots.

Acids, Bases, Buffers and pH

Chemistry | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To measure the pH of strong and weak acids and bases over a range of concentrations, relate dissociation to electrolyte strength, convert between pH and ion concentration, calculate the pH of weak acids and bases from Ka and Kb, and determine the pH of a buffer.

Theory

Strong acids and bases dissociate completely, so pH follows directly from concentration. Weak acids and bases dissociate partially, with the extent set by Ka or Kb. Buffers are governed by the Henderson–Hasselbalch equation.

pH = −log[H⁺] · pOH = −log[OH⁻] · pH + pOH = 14 · Ka = x²/(C−x) · pH = pKa + log([A⁻]/[HA])

Results (selected data)

SolutionC (mol/L)Measured pHTheoretical pH
Strong acid0.012.002.00
Strong acid0.101.001.00
Strong base0.1013.0013.00
Weak acid (Ka 1.8e-5)0.102.88— (weak)
Weak base (Kb 1.8e-5)0.1011.12— (weak)

Worked weak-acid example: Ka = x²/(0.10 − x) = 1.8×10⁻⁵ → x = [H⁺] = 1.33×10⁻³, pH = 2.88, % dissociation = 1.33%.

Buffer example: pKa 4.74, [A⁻] = [HA] = 0.10 → pH = 4.74 + log(1) = 4.74.

Analysis

For the strong acid and strong base the measured pH matched the theoretical values closely (0% deviation in this idealised simulation). As concentration increased ten-fold, the pH of the strong acid fell by one unit each time. Weak acids and bases gave pH values much closer to 7 than strong ones at the same concentration, because only a small fraction dissociates — which also makes them weak electrolytes.

Conclusion

pH depends on both the concentration and the strength (degree of dissociation) of the acid or base. Strong species are strong electrolytes with pH set directly by concentration; weak species are weak electrolytes whose pH is set by Ka or Kb. Buffers hold pH near pKa, shifting only with the ratio of conjugate base to acid.

Practice Questions

Show all work. Use pH = −log[H⁺], pH + pOH = 14, Ka = x²/(C−x), and pH = pKa + log([A⁻]/[HA]).

Question 1
Calculate the pH of 0.001 mol/L HCl (a strong acid).
Hint: [H⁺] = 0.001; pH = −log(0.001) = 3.00.
Question 2
Calculate the pH of 0.01 mol/L NaOH (a strong base).
Hint: pOH = −log(0.01) = 2.00; pH = 14 − 2.00 = 12.00.
Question 3
A solution has pH = 8.5. Find [H⁺] and [OH⁻].
Hint: [H⁺] = 10⁻⁸·⁵; [OH⁻] = 10⁻(14−8.5).
Question 4
A weak acid has Ka = 1.8 × 10⁻⁵ at 0.10 mol/L. Find [H⁺], pH, and percent dissociation.
Hint: solve x²/(0.10−x) = 1.8e-5 → x ≈ 1.33e-3; pH = 2.88; 1.33%.
Question 5
Make a buffer with pKa = 4.74. What ratio of [A⁻]/[HA] gives pH = 5.05?
Hint: 5.05 = 4.74 + log(r) → log(r) = 0.31 → r ≈ 2.0.
Question 6 — Challenge
Explain why a weak acid and a strong acid of the same concentration are both acidic, yet differ greatly in conductivity and pH.
Hint: degree of dissociation — full vs partial — changes free-ion count.