Titration concepts
Define the half-equivalence point.
What is the point in the titration where half of the (NH₃) has been neutralized and converted to its conjugate acid (NH₄⁺)?
At this point, the concentrations of base and conjugate acid are equal. It’s halfway to the stoichiometric (equivalence) point. This is important because it makes pH = pKa (buffer principle).
What does a strong acid vs. weak acid titration curve look like in general?
What is the strong acid curve starts at very low pH with no buffer region, while the weak acid curve starts at higher pH and has a buffer region before the equivalence point?
Strong acids dissociate completely, giving very low initial pH. Weak acids only partially ionize, so they start at higher pH and create a noticeable buffer region before equivalence.
pH after mixing 10.0 mL of 0.10 M HCl with 10.0 mL of 0.10 M NaOH
What is pH = 7
Equal moles neutralize → only water + neutral salt remain.
Initial pH
100.0 mL of 0.050 M NH₃ (Kb = 1.8 × 10⁻⁵) is titrated with 0.100 M HCl.
Calculate the initial pH before any HCl is added.
One Titration Problem (NH₃ titrated with HCl)
Weak base: NH₃
[NH₃]=0.050 M
Kb=1.8×10−5
Volume: 100.0 mL (doesn’t matter here for initial pH)
we assume water is the solvent
equilibrium: NH₃+H₂O↔NH₄⁺+OH⁻
equilibrium expression
Kb=[NH₄⁺][OH]/[NH₃]
Let x=[OH⁻]
Kb=[NH₃]/ x2 ⇒ x=sqr [Kb⋅[NH₃]]
pOH = log (OH-) and 3.52
PH = 14-pOH = 10.48
A buffer is prepared by mixing 0.10 M NH₃ and 0.15 M NH₄Cl. Calculate the pH of the solution.
pOH=pKb+log [NH₄⁺]/[NH₃]
4.74 + log 0.15/0.10
4.74+0.176=4.916
pH=14−4.916 ≈ 9.08
Why is the pH = pKa at the halfway point?
What is because [base] = [acid], so the ratio in the Henderson–Hasselbalch equation is 1, and log(1) = 0, leaving pH = pKa?
Henderson–Hasselbalch: used for acid/base titrations
pH=pKa+log([base][acid])
At half-equivalence, numerator = denominator → ratio = 1 → log(1) = 0 → pH = pKa.
shows that the pH depends on the log of the ratio of base/acid concentration. When this ratio
approaches zero, as it does near the equivalence point, the log function changes dramatically.
When this ratio is much different from zero, as it is well before the equivalence point, the log
function doesn’t change very rapidly.
At the equivalence point of a strong acid–strong base titration, what is the pH and why?
What is pH = 7, because the acid and base completely neutralize each other, leaving only water and neutral ions (like Na⁺ and Cl⁻) in solution?
In SA/SB titrations, neither ion formed (e.g., Na⁺ or Cl⁻) hydrolyzes in water, so the solution is neutral at equivalence. That’s why the equivalence point is always exactly pH 7, unlike weak acid/base titrations.
pH at the equivalence point for SA/SB.
What is 7?
No hydrolysis occurs; neutral solution.
Before Equivalence
After 10.0 mL of HCl has been added, calculate the pH of the solution.
- 100.0 mL of 0.050 M NH₃ with 0.100 M HCl
Determine moles of NH₃ and HCl
nNH₃=M×V=0.050M×0.100L=0.0050mol
nHCl=0.100M×0.010L=0.0010mol
Determine moles of NH₃ and NH₄⁺ after reaction
NH₃+HCl→NH₄⁺+Cl⁻
nNH₃, remaining=0.0050−0.0010=0.0040mol
nNH₄⁺=0.0010mol
V total=100.0mL+10.0mL=110.0mL=0.110L
Convert moles to concentrations
[NH₃]=0.0040/0.110≈ 0.036 M
[NH₄⁺]=0.0010/0.110≈0.00909M
Use the Henderson–Hasselbalch equation - for a weak base with its conjugate acid
pH=14−pOH, pOH=pKb+log [conjugate acid]/[base]
pkb of NH3
pKb= −log(1.8×10−5) ≈ 4.74
pOH= 4.74+log [NH4 +]/[NH₃] = 4.74 + log (0.009/0.036)
pOH = 4.14
pH=14−4.14≈9.86
A buffer is prepared with 0.20 M acetic acid (Ka = 1.8 × 10⁻⁵) and 0.10 M sodium acetate. Calculate the pH.
pH=pKa+log[A⁻]/[HA]=4.74+log0.10/0.20=4.74+ log 0.5
pH=4.74−0.301≈4.44
Why is the equivalence point of a weak base–strong acid titration acidic instead of neutral?
What is because the conjugate acid of the weak base (NH₄⁺, in the case of NH₃) hydrolyzes in water, producing H₃O⁺ and lowering the pH below 7?
In a weak base–strong acid titration, at equivalence all the weak base has been converted to its conjugate acid. That conjugate acid is not neutral — it partially donates protons to water, making the solution acidic. This is why equivalence pH < 7 for WB/SA titrations, unlike SA/SB titrations where equivalence = 7.
Why do weak acids/bases create buffer regions in titrations?
What is because both the weak acid/base and its conjugate are present before equivalence, resisting large pH changes when small amounts of titrant are added?
The coexistence of acid + conjugate base (or base + conjugate acid) creates a buffer. Buffers resist pH change by neutralizing added H⁺ or OH⁻, so the curve has a flat region instead of an abrupt rise.
Draw the titration curve for HCl titrated with NaOH.
What is a curve starting at pH ≈ 1, rising slowly, then a vertical jump at equivalence (pH 7), flattening near 13?
Half-Equivalence Point
At 25.0 mL of HCl added, calculate the pH of the solution.
We are titrating 100.0 mL of 0.050 M NH₃ with 0.100 M HCl, and now 25.0 mL of HCl has been added.
Determine moles of NH₃ and HCl
nNH₃=0.050M×0.100L=0.0050mol
nHCl=0.100M×0.025L=0.0025mol
Determine moles of NH₃ and NH₄⁺ after reaction
NH₃+HCl→NH₄⁺+Cl⁻
nNH₃, remaining=0.0050−0.0025=0.0025mol
nNH₄⁺=0.0025mol
***at half-equivalence, [NH₃]=[NH₄⁺]
Total Volume
Vtotal=100.0mL+25.0mL=125.0mL=0.125L
Concentrations
[NH₃]=0.0025/0.125=0.0200M, [NH₄⁺]=0.0200M
Use the Henderson–Hasselbalch equation
For a weak base buffer:
pOH=pKb+log [conjugate acid]/[base]
[NH₄⁺]=[NH₃] ***
pOH=pKb+log(1)=pKb+0=4.74
pH=14−4.74≈9.26 at half equivalence
Buffer -
A solution like this is called a buffer because it resists changes in pH when small amounts of acid or base are added.
Before equivalence, the solution is a buffer made of NH₃ (base) and NH₄⁺ (conjugate acid).
0.050 M H₂CO₃ (Ka₁ = 4.3×10⁻⁷, Ka₂ = 4.8×10⁻¹¹) is titrated with 0.100 M NaOH. Calculate the pH after adding 25.0 mL NaOH to 50.0 mL of H₂CO₃
We are titrating a diprotic acid (H₂CO₃) with strong base NaOH
Volume H₂CO₃: 50.0 mL = 0.050 L
H₂CO₃ initially: nH₂CO₃=M×V=0.050M×0.050L=0.00250mol
NaOH added: nNaOH=0.100M×0.025L=0.00250mol
Observation: Moles NaOH = moles H₂CO₃ → this is the first equivalence point.
First equivalence point occurs when all first protons (H⁺) from H₂CO₃ are neutralized.
Moles of NaOH needed for first equivalence = moles H₂CO₃ initially = 0.00250 mol.
At the first equivalence point, all H₂CO₃ has been neutralized:H₂CO₃ + NaOH → NaHCO₃ + H₂O
Only HCO₃⁻ remains in solution,
HCO₃⁻ is a conjugate base of weak acid, so the solution is basic
Total volume after titration:
Vtotal=50.0mL H₂CO₃+25.0mL NaOH=75.0mL=0.075L
[HCO₃⁻]=0.00250mol/0.075L≈0.033M
Kb=Kw/Ka2=1.0×10−14/4.8×10−11 ≈ 2.08×10−4 (Determine Kb for HCO₃⁻)
Set up hydrolysis equation: B⁻ + H₂O ⇌ HB + OH⁻
[OH⁻]=Kb⋅[B⁻]=2.08×10−4⋅0.0333 = [OH⁻]≈2.63×10−3M
pOH=−log[OH⁻]=−log(2.63×10−3)≈2.58
pH=14−pOH=14−2.58≈11.42
second equivalence
(No NaOH added)
H₂CO₃ is a weak acid, so we use:[H⁺]=Ka1⋅C=4.3×10−7⋅0.050
pH=−log(1.47×10−4)≈3.83
H ≈ 3.83
Half equivalence point for first proton:
Half-equivalence occurs when half of H₂CO₃ is neutralized → [H₂CO₃] = [HCO₃⁻]
pH = pKa₁ = 6.37
First equivalence point
All H₂CO₃ converted to HCO₃⁻
Moles HCO₃⁻ = 0.00250 mol (from earlier)
Concentration: 0.00250 mol / 0.075 L ≈ 0.0333 M
Kb of HCO₃⁻ = Kw / Ka₂ = 1×10⁻¹⁴ / 4.8×10⁻¹¹ ≈ 2.08×10⁻⁴
pOH = 2.58 → pH = 14 – 2.58 ≈ 11.42
Half - equivalence for second proton
Halfway between first and second equivalence points
At this point: [HCO₃⁻] = [CO₃²⁻]
pH = 14 – pKb (or pKa₂)
pH ≈ 10.32
Second equivalence point
All HCO₃⁻ converted to CO₃²⁻
Concentration: moles = 0.00250 mol, total volume = 50 mL H₂CO₃ + 100 mL NaOH = 150 mL = 0.150 L
[CO₃²⁻]=0.00250/0.150≈0.0167M
CO₃²⁻ is the conjugate base of HCO₃⁻
Kb=Kw/Ka2=2.08×10−4
pOH = -log(1.86×10⁻³) ≈ 2.73 → pH = 14 – 2.73 ≈ 11.27
At the equivalence point of NH₃ + HCl, what species are present in solution?
What are NH₄⁺(aq), Cl⁻(aq), and water (the solution is acidic due to NH₄⁺ hydrolysis)?
At equivalence, all NH₃ is converted into NH₄⁺. Cl⁻ is a spectator ion. NH₄⁺ is a weak acid (conjugate of NH₃) and undergoes hydrolysis, which lowers the pH below 7.
What is the role of an indicator in a titration, and how is it chosen?
What is an indicator changes color near the equivalence point, and it is chosen so its pKa (or color-change range) overlaps with the expected pH at equivalence?
For SA/SB titrations (equivalence ≈ 7), you’d pick an indicator like phenolphthalein (pH 8–10) or bromothymol blue (pH 6–8). For WA/SB or WB/SA titrations, the equivalence point is not at 7, so you must pick accordingly.
Draw the titration curve for a weak acid titrated with a strong base. Be sure to label the initial pH, the buffer region, the equivalence point, and the pH at equivalence.
What is a curve starting above pH 1 (since weak acids don’t fully ionize), showing a buffer region that levels near pH = pKa, a sharp rise at equivalence, and an equivalence point above pH 7 (basic, due to conjugate base hydrolysis)?
Initial pH: higher than a strong acid, because weak acids only partially ionize.
Buffer region: flat region where acid and conjugate base coexist.
Halfway point: pH = pKa.
Equivalence: >7 because the conjugate base (e.g., A⁻) hydrolyzes to make OH⁻.
Curve continues to rise, approaching ~13.
Equivalence Point
At 50.0 mL of HCl added, calculate the pH of the solution.
We are titrating 100.0 mL of 0.050 M NH₃ with 0.100 M HCl, and now 50.0 mL of HCl has been added.
nNH₃=0.050M×0.100L=0.0050mol
nHCl=0.100M×0.050L=0.0050mol
NH₃+HCl→NH₄⁺+Cl⁻
Vtotal=100.0mL+50.0mL=150.0mL=0.150L
[NH₄⁺]=0.0050/0.150≈0.0333M
NH₄⁺ hydrolysis (conjugate acid of a weak base)
At equivalence, the solution contains only NH₄⁺ in water. NH₄⁺ is a weak acid:
NH₄⁺↔NH₃+H⁺
The acid dissociation constant of NH₄⁺:
Ka=Kw/Kb=1.0×10−14/1.8×10−5 ≈ 5.56×10−10
[H⁺]= sqr Ka⋅[NH₄⁺]=(5.56×10−10)(0.0333)
pH=−log[H⁺]=−log(4.30×10−6)≈5.37
Notice: pH < 7 because the conjugate acid of a weak base (NH₄⁺) makes the solution slightly acidic
Why are titrations useful in real-world settings (e.g., industry, medicine, environment)?
What is they allow quantitative determination of concentration and purity—for example, in pharmaceuticals, water quality testing, and food industry acid/base balance?
Titrations are not just a lab exercise. In clinical labs, they check blood acidity; in industry, they measure acidity/basicity in food, beverages, and chemicals; in environmental testing, they monitor pollutants. The method is precise, inexpensive, and widely used.
Compare titrations of polyprotic acids vs. monoprotic acids. What’s different about their curves?
What is polyprotic acids show multiple equivalence points (one per proton lost), while monoprotic acids have a single equivalence point?
For example, H₂SO₄ has 2 protons → 2 equivalence points; H₃PO₄ has 3 → 3 equivalence points. Each proton comes off stepwise, and the titration curve shows multiple “jumps” in pH. A monoprotic acid like HCl only has one.
Draw the titration curve for a weak base titrated with a strong acid. Be sure to label the starting pH, the buffer region, the half-equivalence point (pH = pKa), and the equivalence point (acidic).
What is a curve that starts in the basic range (~10–11), shows a buffer region, drops steeply near equivalence, has pH < 7 at equivalence (due to conjugate acid hydrolysis), and then levels off near pH 1–2?
Start: Weak base (NH₃) solution is basic, but not as high as strong bases.
Buffer: Region where NH₃/NH₄⁺ coexist.
Half-equivalence: pH = pKa (~9.25 for NH₄⁺).
Equivalence: acidic (pH ≈ 5–6).
After equivalence: excess strong acid dominates, pH ~1–2.
Titrate 100.0 mL of 0.050 M NH₃ with 0.100 M HCl.
Draw a pH vs. volume of HCl added graph, and label: initial pH, half-equivalence point, equivalence point, and any relevant buffer regions.
Mark key volumes on the x-axis (volume of HCl added):
0 mL → initial pH (calculated earlier, ~10.48)
25.0 mL → half-equivalence point (pH ~9.26)
50.0 mL → equivalence point (pH ~5.37)
>50 mL → after equivalence (pH drops more steeply toward strong acid, below 5.37)
Draw the curve:
Start basic (>7) because NH₃ is a weak base.
Slowly decrease as HCl is added, showing the buffer region (relatively gradual slope) between 0 mL and ~50 mL.
Half-equivalence is roughly the middle of the buffer region — mark it.
Equivalence point shows the steepest drop in pH, because all NH₃ is neutralized.
After equivalence, pH falls more slowly and approaches the pH of the excess strong acid.
Optional: Label the buffer region clearly.
For a weak base/strong acid titration, the buffer exists from roughly 0 to just before equivalence