UNIT 8

Acids and Bases

Explore acid–base definitions, pH calculations, weak-acid and weak-base equilibria, buffers, titrations, molecular structure, and indicators.

Topics

Acid–Base Fundamentals and pH

Identify acids, bases, conjugate pairs, and amphoteric substances and calculate pH and pOH for strong acids and bases.

Weak Acids, Weak Bases, and Buffers

Use Ka, Kb, and equilibrium calculations to analyze weak acids, weak bases, conjugates, and buffer solutions.

Titrations and Indicators

Analyze acid–base titrations, equivalence points, titration curves, buffer regions, and indicator selection.

Acid–Base Fundamentals and pH

Acids and bases can be identified by how they behave in water and how they transfer protons during chemical reactions.

Arrhenius Acids and Bases

The Arrhenius definition focuses on substances dissolved in water.

Arrhenius acid: Produces H+, or H3O+, in water.

Arrhenius base: Produces OH− in water.

For example:

HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)
NaOH(aq) → Na+(aq) + OH−(aq)

Brønsted–Lowry Acids and Bases

The Brønsted–Lowry definition describes acid–base reactions as transfers of H+, also called protons.

Brønsted–Lowry acid: Proton donor

Brønsted–Lowry base: Proton acceptor

Example: Proton Transfer

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)

NH3 accepts H+, so it acts as a base. H2O donates H+, so it acts as an acid.

Conjugate Acid–Base Pairs

When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

Acid ⇌ H+ + Conjugate base
Base + H+ ⇌ Conjugate acid

Consider the reaction:

HCl + H2O → H3O+ + Cl−
Species Role Conjugate Partner
HCl Acid Cl−
H2O Base H3O+
Members of a conjugate acid–base pair differ by exactly one H+.

Amphiprotic Substances

An amphiprotic substance can either donate or accept a proton, depending on the substance with which it reacts.

Water is amphiprotic. It can accept H+ to form H3O+, or donate H+ to form OH−.

HCO3− is another amphiprotic species:

HCO3− + H+ → H2CO3
HCO3− → CO32− + H+

Strong and Weak Acids

A strong acid ionizes essentially completely in water. A weak acid ionizes only partially and establishes an equilibrium.

Strong acid: HA + H2O → H3O+ + A−
Weak acid: HA + H2O ⇌ H3O+ + A−

Common strong acids include:

Strong and Weak Bases

Strong bases dissociate essentially completely in water. Common strong bases include soluble Group 1 metal hydroxides and the more soluble Group 2 metal hydroxides.

NaOH(aq) → Na+(aq) + OH−(aq)

Weak bases react only partially with water.

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
Acid or base strength is not the same as concentration.

Strength describes the extent of ionization. Concentration describes how much solute is present per volume of solution.

Autoionization of Water

A small fraction of water molecules transfer protons to one another.

2H2O(l) ⇌ H3O+(aq) + OH−(aq)

The equilibrium constant for this process is called the ion-product constant for water, Kw.

Kw = [H3O+][OH−]

At 25°C:

Kw = 1.0 × 10−14

In neutral water at 25°C:

[H3O+] = [OH−] = 1.0 × 10−7 M

The pH Scale

pH scale from zero to fourteen showing acidic, neutral, and basic regions
At 25°C, solutions with pH below 7 are acidic, pH 7 is neutral, and pH above 7 is basic.
pH = −log[H3O+]
pOH = −log[OH−]

At 25°C:

pH + pOH = 14.00
The pH scale is logarithmic. A decrease of one pH unit corresponds to a tenfold increase in [H3O+].

Although the commonly displayed pH scale runs from 0 to 14, sufficiently concentrated solutions can have pH values outside this range.

Calculating pH of a Strong Acid

Example: HCl Solution

Calculate the pH of 0.00320 M HCl.

HCl is a strong acid and ionizes completely. Because it supplies one H3O+ for each HCl:

[H3O+] = 0.00320 M
pH = −log(0.00320)
pH = 2.495

Calculating pH of a Strong Base

Example: Ba(OH)2 Solution

Calculate the pH of 0.0150 M Ba(OH)2 at 25°C.

Each formula unit produces two hydroxide ions:

Ba(OH)2 → Ba2+ + 2OH−
[OH−] = 2(0.0150 M) = 0.0300 M
pOH = −log(0.0300) = 1.523
pH = 14.000 − 1.523 = 12.477

Finding Concentration from pH

The inverse logarithmic relationships are:

[H3O+] = 10−pH
[OH−] = 10−pOH

Example: Finding Hydronium Concentration

Calculate [H3O+] in a solution with a pH of 4.35.

[H3O+] = 10−4.35
[H3O+] = 4.5 × 10−5 M

Practice Questions

Question 1: Conjugate Acid–Base Pairs

Consider the reaction:

H2PO4− + H2O ⇌ H3O+ + HPO42−

Identify the acid, base, conjugate acid, and conjugate base.

Show solution

H2PO4− donates H+, so it is the acid.

H2O accepts H+, so it is the base.

HPO42− is the conjugate base of H2PO4−.

H3O+ is the conjugate acid of H2O.

Question 2: Amphiprotic Behavior

Explain why HSO4− can be classified as amphiprotic.

Show solution

HSO4− can donate a proton and become SO42−:

HSO4− → H+ + SO42−

It can also accept a proton and become H2SO4:

HSO4− + H+ → H2SO4

Because it can either donate or accept a proton, it is amphiprotic.

Question 3: Calculating Strong-Acid pH

Calculate the pH of a 0.00250 M HNO3 solution.

Show solution

HNO3 is a strong acid and ionizes essentially completely.

[H3O+] = 0.00250 M
pH = −log(0.00250)
pH = 2.602

Question 4: A Strong Base with Two Hydroxides

Calculate the pH of 0.0125 M Ca(OH)2 at 25°C. Assume complete dissociation.

Show solution
Ca(OH)2 → Ca2+ + 2OH−

Each mole of Ca(OH)2 produces two moles of OH−.

[OH−] = 2(0.0125 M) = 0.0250 M
pOH = −log(0.0250) = 1.602
pH = 14.000 − 1.602
pH = 12.398

Question 5: Concentrations from pH

A solution has a pH of 3.72 at 25°C. Calculate [H3O+] and [OH−].

Show solution

Calculate the hydronium concentration:

[H3O+] = 10−3.72
[H3O+] = 1.9 × 10−4 M

Use Kw to find hydroxide concentration:

[OH−] = Kw [H3O+]
[OH−] = 1.0 × 10−14 1.9 × 10−4
[OH−] = 5.3 × 10−11 M

Question 6: Comparing pH Values

Solution X has a pH of 2.50, while Solution Y has a pH of 4.50. How many times greater is [H3O+] in Solution X?

Show solution

The difference in pH is:

4.50 − 2.50 = 2.00

Each pH unit represents a factor of ten in hydronium concentration.

102.00 = 100

Solution X has 100 times greater [H3O+] than Solution Y.

Question 7: Strength Versus Concentration

A student claims that a concentrated weak acid must be stronger than a dilute strong acid. Explain why this statement is incorrect.

Show solution

Strength and concentration describe different properties.

Acid strength describes the extent to which an acid ionizes. A strong acid ionizes essentially completely, while a weak acid ionizes only partially.

Concentration describes the amount of acid present per volume of solution. A weak acid can be concentrated, and a strong acid can be dilute, without changing their classifications.

VIDEO EXPLANATION

Watch: Strong Acids, Bases, and pH

Watch a worked example showing how to identify conjugate pairs and calculate pH, pOH, and ion concentrations.

Watch on TikTok →

Weak Acids, Weak Bases, and Buffers

Weak acids and bases react only partially with water. Their solutions must therefore be analyzed as equilibrium systems rather than by assuming complete ionization.

Weak-Acid Equilibrium

A weak acid, represented by HA, transfers only a fraction of its protons to water.

HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq)

The acid-ionization constant is:

Ka = [H3O+][A−] [HA]
A larger Ka indicates a stronger weak acid and a greater extent of ionization.

A smaller Ka indicates a weaker acid and a smaller extent of ionization.

Calculating the pH of a Weak Acid

Example: Weak-Acid ICE Table

Calculate the pH of 0.100 M CH3COOH. Ka = 1.8 × 10−5.

CH3COOH + H2O ⇌ H3O+ + CH3COO−
CH3COOH H3O+ CH3COO−
Initial 0.100 0 0
Change −x +x +x
Equilibrium 0.100 − x x x
1.8 × 10−5 = x2 0.100 − x

Because Ka is small, x is expected to be much smaller than 0.100. Approximate 0.100 − x as 0.100.

1.8 × 10−5 ≈ x2 0.100
x = 1.34 × 10−3 M
[H3O+] = 1.34 × 10−3 M
pH = −log(1.34 × 10−3) = 2.87

The Five-Percent Check

When x is removed from an initial concentration, the approximation that initial concentration minus x is nearly equal to the initial concentration should be checked.

Percent change = x Initial concentration × 100

For the previous example:

Percent change = 0.00134 0.100 × 100 = 1.34%

Because the result is less than 5%, the approximation is reasonable. If it exceeds 5%, solve the complete equation, usually with the quadratic formula.

Percent Ionization

Percent ionization = [H3O+]equilibrium [HA]initial × 100
Diluting a weak acid increases its percent ionization, even though its hydronium concentration and acidity decrease.

Weak-Base Equilibrium

A weak base accepts a proton from water and produces its conjugate acid and hydroxide.

B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq)
Kb = [BH+][OH−] [B]

Example: Weak-Base pH

Calculate the pH of 0.200 M NH3. Kb = 1.8 × 10−5.

NH3 + H2O ⇌ NH4+ + OH−
1.8 × 10−5 ≈ x2 0.200
x = [OH−] = 1.90 × 10−3 M
pOH = −log(1.90 × 10−3) = 2.72
pH = 14.00 − 2.72 = 11.28

Conjugate Strength

The stronger an acid is, the weaker its conjugate base is. Similarly, the stronger a base is, the weaker its conjugate acid is.

KaKb = Kw

For a conjugate acid–base pair at 25°C:

KaKb = 1.0 × 10−14

Example: Finding a Conjugate Kb

An acid has Ka = 2.0 × 10−5. Find Kb for its conjugate base.

Kb = Kw Ka
Kb = 1.0 × 10−14 2.0 × 10−5
Kb = 5.0 × 10−10

pKa and Acid Strength

pKa = −log(Ka)
A larger Ka means a stronger acid.

A smaller pKa means a stronger acid.

Molecular Structure and Acid Strength

Acid strength depends on how easily a molecule can lose H+ and how stable its conjugate base is.

For oxyacids with the same central atom, additional oxygen atoms generally increase acid strength because they withdraw electron density and stabilize the conjugate base.

HClO < HClO2 < HClO3 < HClO4

Polyprotic Acids

A polyprotic acid can donate more than one proton. Each proton is removed in a separate equilibrium step.

For a diprotic acid H2A:

H2A + H2O ⇌ H3O+ + HA−
HA− + H2O ⇌ H3O+ + A2−
Successive ionizations become less favorable:

Ka1 > Ka2 > Ka3

Buffer Solutions

A buffer resists large changes in pH when a small amount of strong acid or strong base is added.

A buffer commonly contains either:

Example: How a Buffer Responds

Consider a buffer containing HA and A−.

When strong acid is added:

A− + H3O+ → HA + H2O

When strong base is added:

HA + OH− → A− + H2O

The buffer converts the added strong acid or base into a weaker species, reducing the resulting change in pH.

The Henderson–Hasselbalch Equation

The pH of a weak-acid buffer can be calculated using:

pH = pKa + log [A−] [HA]
When [A−] = [HA], the logarithm equals zero and:

pH = pKa

Example: Calculating Buffer pH

A buffer contains 0.100 M HA and 0.200 M A−. The acid has a pKa of 4.74. Calculate the pH.

pH = 4.74 + log 0.200 0.100
pH = 4.74 + log(2.00)
pH = 5.04

Buffer Capacity

Buffer capacity describes how much strong acid or strong base a buffer can absorb before its pH changes substantially.

Buffer capacity increases when the concentrations of both buffer components increase.

A buffer works most effectively when the weak acid and conjugate base concentrations are similar.

Henderson–Hasselbalch is generally most useful when the ratio of conjugate base to weak acid is between approximately 0.1 and 10.

Adding Strong Acid or Base to a Buffer

When a strong acid or base is added to a buffer, perform the stoichiometric reaction first. Then use the remaining amounts of weak acid and conjugate base in the Henderson–Hasselbalch equation.

Buffer problem order:

1. React the added strong acid or base completely

2. Determine the new amounts of HA and A−

3. Use Henderson–Hasselbalch to calculate the new pH

Practice Questions

Question 1: Weak-Acid pH

Calculate the pH of a 0.200 M solution of a weak acid HA with Ka = 4.0 × 10−6.

Show solution
HA + H2O ⇌ H3O+ + A−
HA H3O+ A−
Initial 0.200 0 0
Change −x +x +x
Equilibrium 0.200 − x x x
4.0 × 10−6 ≈ x2 0.200
x = 8.9 × 10−4 M
pH = −log(8.9 × 10−4) = 3.05

The solution has a pH of 3.05.

Question 2: Percent Ionization

Using the results from Question 1, calculate the percent ionization of the weak acid.

Show solution
Percent ionization = [H3O+]eq [HA]initial × 100
Percent ionization = 8.9 × 10−4 0.200 × 100
Percent ionization = 0.45%

Because this result is below 5%, the approximation used in Question 1 is reasonable.

Question 3: Weak-Base pH

Calculate the pH of 0.150 M weak base B if Kb = 2.5 × 10−5.

Show solution
B + H2O ⇌ BH+ + OH−
2.5 × 10−5 ≈ x2 0.150
x = [OH−] = 1.94 × 10−3 M
pOH = −log(1.94 × 10−3) = 2.71
pH = 14.00 − 2.71 = 11.29

Question 4: Conjugate-Base Strength

A weak acid has Ka = 6.3 × 10−5. Calculate Kb for its conjugate base at 25°C.

Show solution
KaKb = Kw
Kb = 1.0 × 10−14 6.3 × 10−5
Kb = 1.6 × 10−10

The relatively small Kb is consistent with the inverse relationship between an acid’s strength and the strength of its conjugate base.

Question 5: Comparing Weak-Acid Strength

Three weak acids have the following values:

  • Acid X: Ka = 1.0 × 10−3
  • Acid Y: Ka = 4.0 × 10−6
  • Acid Z: Ka = 7.0 × 10−9

Rank the acids from strongest to weakest. Then rank their conjugate bases from strongest to weakest.

Show solution

A larger Ka indicates a stronger acid.

Acid strength: X > Y > Z

Conjugate-base strength follows the reverse order.

Conjugate-base strength: Z− > Y− > X−

Question 6: Buffer pH

A buffer contains 0.150 M weak acid HA and 0.300 M conjugate base A−. The acid has a pKa of 4.20.

Calculate the buffer’s pH.

Show solution
pH = pKa + log [A−] [HA]
pH = 4.20 + log 0.300 0.150
pH = 4.20 + log(2.00)
pH = 4.50

Question 7: Adding Strong Acid to a Buffer

A buffer initially contains 0.100 mol HA and 0.150 mol A−. The acid has pKa = 4.76.

Calculate the pH after 0.0200 mol HCl is added. Assume the volume change is negligible.

Show solution

The added strong acid reacts with the conjugate base:

A− + H3O+ → HA + H2O

Subtract the added acid from A− and add the same amount to HA.

A−: 0.150 − 0.0200 = 0.130 mol
HA: 0.100 + 0.0200 = 0.120 mol

Because both substances occupy the same final volume, their mole ratio can be used directly.

pH = 4.76 + log 0.130 0.120
pH = 4.79

Question 8: Selecting a Buffer

Which mixture can function as a buffer? Explain.

  • HCl and NaCl
  • HF and NaF
  • NaOH and NaCl
Show solution

HF and NaF can form a buffer.

HF is a weak acid, and NaF supplies its conjugate base, F−. Both components are available to react with small additions of strong base or strong acid.

HCl and NaOH are strong, so the other listed mixtures do not contain the required weak species and its conjugate partner.

Question 9: Dilution and Percent Ionization

A weak acid solution is diluted with water. Explain what happens to its hydronium concentration, pH, and percent ionization.

Show solution

Dilution decreases [H3O+], so the solution’s pH increases.

However, dilution also shifts the weak-acid ionization equilibrium toward the ions. Therefore, a greater percentage of the acid molecules ionize.

The hydronium concentration decreases while the percent ionization increases.

VIDEO EXPLANATION

Watch: Weak Acids and Buffers

Watch a worked example involving weak-acid equilibrium, percent ionization, conjugate strength, or buffer pH.

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Titrations and Indicators

An acid–base titration uses a solution with a known concentration to determine the amount or concentration of an acid or base in another solution.

Important Titration Terms

Term Meaning
Titrant The solution of known concentration that is gradually added
Analyte The solution whose concentration or amount is being determined
Equivalence point The point at which stoichiometrically equivalent amounts of acid and base have reacted
Endpoint The observed indicator color change used to estimate the equivalence point
The equivalence point is determined by reaction stoichiometry. The endpoint is an experimental observation. A suitable indicator makes the endpoint occur close to the equivalence point.

Titration Stoichiometry

Before calculating pH, first determine the number of moles of acid and base that react.

Moles = Molarity × Volume in liters

Use the balanced neutralization equation to compare the acid and base amounts.

Example: Finding an Unknown Concentration

A 25.00 mL HCl sample is neutralized by 20.00 mL of 0.1500 M NaOH. Calculate the HCl concentration.

HCl + NaOH → NaCl + H2O

Calculate the moles of NaOH:

nNaOH = (0.1500 mol/L)(0.02000 L) = 0.003000 mol

HCl and NaOH react in a one-to-one ratio, so the sample contains 0.003000 mol HCl.

MHCl = 0.003000 mol 0.02500 L
MHCl = 0.1200 M

Titration Curves

A titration curve displays pH on the vertical axis and volume of titrant added on the horizontal axis.

Strong acid strong base and weak acid strong base titration curves with important regions labeled
A strong acid–strong base titration has an equivalence point near pH 7 at 25°C. A weak acid–strong base titration has a buffer region, a half-equivalence point, and an equivalence point above pH 7.

Strong Acid–Strong Base Titration

Consider a strong acid being titrated with a strong base. Both substances ionize essentially completely.

H3O+ + OH− → 2H2O
Region How pH Is Determined
Before titrant is added Initial strong-acid concentration
Before equivalence Excess moles of strong acid
At equivalence Approximately pH 7 at 25°C
After equivalence Excess moles of strong base
The pH at the equivalence point is 7 only for a strong acid–strong base titration at 25°C. Not every equivalence point occurs at pH 7.

Calculating pH Before Equivalence

Example: Excess Strong Acid

A 25.0 mL sample of 0.100 M HCl is titrated with 15.0 mL of 0.100 M NaOH. Calculate the pH.

Calculate the initial moles of HCl:

nHCl = (0.100 M)(0.0250 L) = 0.00250 mol

Calculate the moles of NaOH added:

nNaOH = (0.100 M)(0.0150 L) = 0.00150 mol

Determine the excess HCl:

0.00250 − 0.00150 = 0.00100 mol HCl

Divide by the total solution volume:

[H3O+] = 0.00100 mol 0.0400 L = 0.0250 M
pH = −log(0.0250) = 1.60

Weak Acid–Strong Base Titration

A weak acid–strong base titration contains several chemically different regions.

Region Major Species or Calculation
Initial solution Weak-acid equilibrium using Ka
Before equivalence Buffer containing HA and A−
Half-equivalence point [HA] = [A−] and pH = pKa
Equivalence point Conjugate base hydrolysis; pH greater than 7
After equivalence Excess strong base determines pH

The Half-Equivalence Point

At the half-equivalence point, exactly half of the original weak acid has been converted into its conjugate base.

[HA] = [A−]
pH = pKa

This point can be used to determine Ka from an experimental titration curve.

Example: Finding Ka from a Curve

A weak acid reaches its equivalence point after 30.0 mL of strong base has been added. The pH after 15.0 mL has been added is 4.60.

Because 15.0 mL is half of the equivalence-point volume, this is the half-equivalence point.

pKa = 4.60
Ka = 10−4.60
Ka = 2.5 × 10−5

Why the Weak-Acid Equivalence Point Is Basic

At equivalence, the weak acid has been converted into its conjugate base. That conjugate base reacts with water to produce OH−.

A− + H2O ⇌ HA + OH−

Therefore, the equivalence-point pH is greater than 7 for a weak acid titrated with a strong base.

Strong Acid–Weak Base Titrations

If a weak base is titrated with a strong acid, the equivalence solution contains the weak base’s conjugate acid.

BH+ + H2O ⇌ B + H3O+

The conjugate acid produces hydronium, so the equivalence-point pH is less than 7.

Comparing Equivalence Points

Titration Equivalence-Point pH at 25°C
Strong acid with strong base Approximately 7
Weak acid with strong base Greater than 7
Weak base with strong acid Less than 7

Acid–Base Indicators

An acid–base indicator is a weak acid or weak base whose two forms have different colors.

HIn ⇌ H+ + In−

The color transition occurs over a range of pH values near the indicator’s pKa.

Choose an indicator whose color-change range lies within the steep vertical region of the titration curve.

An indicator does not need to change color exactly at pH 7. Its useful range must instead overlap the rapid pH change near the equivalence point.

Common Titration Strategy

1. Calculate moles of acid and base

2. Complete the neutralization reaction

3. Identify which titration region applies

4. Use the correct remaining species to calculate pH

Practice Questions

Question 1: Identifying a Titration Curve

A titration curve begins at a moderately acidic pH, contains a buffer region, and has an equivalence point above pH 7.

What type of titration does the curve represent?

Show solution

The curve represents a weak acid titrated with a strong base.

The weak acid causes the initial pH to be higher than that of a strong acid with the same concentration. The mixture of weak acid and conjugate base creates the buffer region.

At equivalence, the conjugate base reacts with water to produce OH−, causing the pH to be greater than 7.

Question 2: Diprotic-Acid Stoichiometry

A 20.00 mL sample of H2SO4 is neutralized by 30.00 mL of 0.1600 M NaOH.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Calculate the concentration of the H2SO4 solution.

Show solution

Calculate the moles of NaOH:

nNaOH = (0.1600 mol/L)(0.03000 L) = 0.004800 mol

Two moles of NaOH react with one mole of H2SO4.

nH₂SO₄ = 0.004800 mol NaOH × 1 mol H2SO4 2 mol NaOH
nH₂SO₄ = 0.002400 mol
MH₂SO₄ = 0.002400 mol 0.02000 L
MH₂SO₄ = 0.1200 M

Question 3: pH after the Equivalence Point

A 25.0 mL sample of 0.100 M HCl is titrated with 30.0 mL of 0.100 M NaOH. Calculate the pH after the base has been added.

Show solution

Calculate the initial moles of acid:

nHCl = (0.100 M)(0.0250 L) = 0.00250 mol

Calculate the moles of base added:

nNaOH = (0.100 M)(0.0300 L) = 0.00300 mol

Determine the excess hydroxide:

Excess OH− = 0.00300 − 0.00250 = 0.00050 mol

The total volume is:

25.0 mL + 30.0 mL = 55.0 mL = 0.0550 L
[OH−] = 0.00050 mol 0.0550 L = 0.00909 M
pOH = −log(0.00909) = 2.04
pH = 14.00 − 2.04 = 11.96

Question 4: Half-Equivalence Point

A weak acid reaches its equivalence point after 40.0 mL of strong base is added. The pH after 20.0 mL of base has been added is 5.20.

Determine the acid’s pKa and Ka.

Show solution

Half of 40.0 mL is 20.0 mL, so this measurement was taken at the half-equivalence point.

pH = pKa = 5.20
Ka = 10−5.20
Ka = 6.3 × 10−6

Question 5: pH in the Buffer Region

A sample initially contains 0.00250 mol of weak acid HA. During its titration, 0.00100 mol of strong base is added. The acid has a pKa of 4.74.

Calculate the pH before the equivalence point. Assume the base reacts completely with HA.

Show solution
HA + OH− → A− + H2O

The added base consumes 0.00100 mol HA and produces 0.00100 mol A−.

Remaining HA = 0.00250 − 0.00100 = 0.00150 mol
A− formed = 0.00100 mol

Use Henderson–Hasselbalch. The mole ratio can be used because both species occupy the same volume.

pH = 4.74 + log 0.00100 0.00150
pH = 4.74 + log(0.667)
pH = 4.56

Question 6: Equivalence-Point Reasoning

Explain why the equivalence-point pH is above 7 when a weak acid is titrated with a strong base.

Show solution

At equivalence, the strong base has converted the weak acid into its conjugate base.

A− + H2O ⇌ HA + OH−

The conjugate base reacts with water and produces OH−. Therefore, the equivalence-point solution is basic and has a pH above 7.

Question 7: Selecting an Indicator

A titration has an equivalence point near pH 8.8. Three indicators have the following transition ranges:

  • Indicator A: pH 3.1–4.4
  • Indicator B: pH 6.0–7.6
  • Indicator C: pH 8.2–10.0

Which indicator is the best choice?

Show solution

Indicator C is the best choice because its transition range includes pH 8.8.

Its color change should occur within the steep region surrounding the equivalence point.

Question 8: Titration Error Analysis

A student titrates an unknown acid but adds excess base beyond the indicator endpoint. The student uses the recorded volume to calculate the acid concentration.

Will the calculated acid concentration be too high or too low? Explain.

Show solution

The calculated acid concentration will be too high.

The recorded base volume is greater than the volume actually required for neutralization. This causes the calculated moles of base—and therefore the calculated moles of acid—to be too large.

Question 9: Comparing Titration Curves

Equal concentrations and volumes of a strong acid and weak acid are separately titrated with the same strong base. Compare their initial pH values and equivalence-point volumes.

Show solution

The strong acid has the lower initial pH because it ionizes essentially completely.

The weak acid has the higher initial pH because it ionizes only partially.

The equivalence-point volumes are the same because the samples contain the same total moles of monoprotic acid. Acid strength changes the initial pH, but not the number of moles requiring neutralization.

VIDEO EXPLANATION

Watch: Titration Curves

Watch a worked example showing how to identify titration regions, calculate pH, locate the half-equivalence point, and select an indicator.

Watch on TikTok →