Introduction to Titration滴定入门

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 4~50 min

Learning goals

  • Define titrant, analyte, equivalence point, and end point, and identify the equivalence point from the amounts of titrant and analyte
  • Use titration data to calculate the concentration of a solution or the mass of a substance, for acid–base titrations and for redox titrations with KMnO₄
  • Analyze how titration technique (rinsing, reading the buret, discarding data, primary standards) affects the result

Before you start

  • Neutralization and titration (2.12.5)
  • Solution stoichiometry (2.9.4)
  • Stoichiometry (U4.5)

Using a known to measure an unknown

Lesson 2.12.5 introduced acid–base titration. AP topic 4.6 treats titration as a stoichiometric method: a solution of accurately known concentration is used to measure the amount of another substance.

  • The titrant滴定剂 is the solution of accurately known concentration (a standard solution) in the buret.
  • The analyte待测物 is the substance being measured, in the flask.
  • At the equivalence point化学计量点, the titrant added has reacted exactly with the analyte in the ratio given by the balanced equation.
  • The end point滴定终点 is where the indicator changes color and the titration stops. A suitable indicator makes the end point as close to the equivalence point as possible.

Finding the equivalence point

At the equivalence point, the moles of titrant and analyte are in the coefficient ratio. For a monoprotic acid reacting 1 : 1 with NaOH:

MacidVacid=MbaseVbaseM_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}

The titration reaction must go essentially to completion for the equivalence point to be calculated accurately from moles.

Strong and weak acids: the same equivalence volume

Equal moles of a monoprotic strong acid and a monoprotic weak acid need the same amount of NaOH, because OH⁻ also reacts essentially completely with weak-acid molecules. What differs is the pH at the equivalence point: 7 for the strong acid, above 7 for the weak acid (the CH₃COO⁻ formed is a weak base).

Halfway to the equivalence point, a weak acid solution contains equal amounts of CH₃COOH and CH₃COO⁻, and the pH is about equal to pKₐ. Drag the slider to watch the particles in the flask change:

Try it: which particles are in the flask during a titration?
H⁺H⁺H⁺H⁺Cl⁻Cl⁻Cl⁻Cl⁻H⁺H⁺H⁺Cl⁻Cl⁻Cl⁻Cl⁻Na⁺H⁺H⁺Cl⁻Cl⁻Cl⁻Cl⁻Na⁺Na⁺H⁺Cl⁻Cl⁻Cl⁻Cl⁻Na⁺Na⁺Na⁺Cl⁻Cl⁻Cl⁻Cl⁻Na⁺Na⁺Na⁺Na⁺Cl⁻Cl⁻Cl⁻Cl⁻Na⁺Na⁺Na⁺Na⁺Na⁺OH⁻Cl⁻Cl⁻Cl⁻Cl⁻Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺OH⁻OH⁻CH₃COOHCH₃COOHCH₃COOHCH₃COOHCH₃COOHCH₃COOHCH₃COOHCH₃COO⁻Na⁺CH₃COOHCH₃COOHCH₃COO⁻CH₃COO⁻Na⁺Na⁺CH₃COOHCH₃COO⁻CH₃COO⁻CH₃COO⁻Na⁺Na⁺Na⁺CH₃COO⁻CH₃COO⁻CH₃COO⁻CH₃COO⁻Na⁺Na⁺Na⁺Na⁺CH₃COO⁻CH₃COO⁻CH₃COO⁻CH₃COO⁻Na⁺Na⁺Na⁺Na⁺Na⁺OH⁻CH₃COO⁻CH₃COO⁻CH₃COO⁻CH₃COO⁻Na⁺Na⁺Na⁺Na⁺Na⁺Na⁺OH⁻OH⁻Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe²⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Mn²⁺K⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Mn²⁺Mn²⁺K⁺K⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Fe³⁺Mn²⁺Mn²⁺MnO₄⁻K⁺K⁺K⁺

The titration curves show the same thing: both jumps come at about 25.00 mL.

Try it: titrate 25.00 mL of 0.100 M acid with 0.100 M NaOH
Acid
Indicator
mL

    Redox titrations

    A titration reaction doesn’t have to be acid–base. In acidic solution, Fe²⁺ can be titrated with KMnO₄:

    MnOX4X−(aq)+5 FeX2+(aq)+8 HX+(aq)→MnX2+(aq)+5 FeX3+(aq)+4 HX2O(l)\ce{MnO4-(aq) + 5Fe^2+(aq) + 8H+(aq) -> Mn^2+(aq) + 5Fe^3+(aq) + 4H2O(l)}
    • MnO₄⁻ and Fe²⁺ react in a 1 : 5 ratio.
    • No added indicator is needed. MnO₄⁻ is purple and Mn²⁺ is nearly colorless. Before the equivalence point, each drop of MnO₄⁻ loses its color at once; after the Fe²⁺ is used up, one more drop turns the solution pale pink: the end point.
    • The reaction uses up H⁺, so the pH gradually rises.

    KMnO₄ can also titrate H₂O₂: 2 MnOX4X−(aq)+5 HX2OX2(aq)+6 HX+(aq)→2 MnX2+(aq)+5 OX2(g)+8 HX2O(l)\ce{2MnO4-(aq) + 5H2O2(aq) + 6H+(aq) -> 2Mn^2+(aq) + 5O2(g) + 8H2O(l)}, a 2 : 5 ratio.

    Technique and errors

    • Buret: after washing with distilled water, rinse it two or three times with the titrant. Otherwise leftover water dilutes the titrant, more volume is needed, and the calculated analyte concentration is too high.
    • Flask: it doesn’t need to be dry, and it must not be rinsed with the analyte. Water in the flask doesn’t change the moles of analyte; rinsing with analyte adds more.
    • Reading: read the bottom of the meniscus at eye level, to 0.01 mL.
    • Discarding data: run several trials and discard one that is clearly out of line with the others (often a misread buret or an overshot end point) before averaging.
    • Primary standards: NaOH solutions are often standardized with potassium hydrogen phthalate (KHP, KHC₈H₄O₄). If the KHP is wet, part of the weighed mass is water, so there’s less KHP than calculated; less NaOH is needed, and the calculated NaOH concentration is too high.
    • Measuring a gas volume: collect the gas in a water-filled eudiometer (gas-measuring tube) inverted in water, and read the gas volume directly.

    Example 1: Standardizing NaOH with KHP

    0.5105 g of KHP (molar mass 204.22 g/mol) is dissolved in water and titrated with NaOH solution; the end point comes at 25.00 mL. KHP and NaOH react 1 : 1. Find the molarity of the NaOH. How would wet KHP affect the result?

    Step-by-step solution

    1. Moles of KHP. KHP is the weighed primary standard, so its moles come from its mass (in mol):

      n(KHP)=0.5105204.22=2.500×10−3\begin{aligned} n(\text{KHP}) &= \frac{0.5105}{204.22} \\ &= 2.500 \times 10^{-3} \end{aligned}

    2. Moles of NaOH from the ratio. At the end point they have reacted exactly 1 : 1:

      n(NaOH)=n(KHP)=2.500×10−3\begin{aligned} n(\ce{NaOH}) &= n(\text{KHP}) \\ &= 2.500 \times 10^{-3} \end{aligned}

    3. Molarity of NaOH. Convert the volume to liters: 25.00 mL = 0.02500 L:

      [NaOH]=2.500×10−30.02500=0.1000 M\begin{aligned} [\ce{NaOH}] &= \frac{2.500 \times 10^{-3}}{0.02500} \\ &= \boxed{0.1000\ \text{M}} \end{aligned}

    4. Error analysis: wet KHP. First, which quantity is off? Part of the 0.5105 g is water, so the actual KHP is less than 2.500 × 10⁻³ mol and the NaOH volume V used is smaller. Then, where does it sit in the formula? The calculation still uses 0.5105 g (numerator unchanged) while the smaller V is in the denominator, so the calculated molarity is too high.

    5. Check significant figures. 0.5105 and 25.00 both have 4, so the answer has 4.

    Example 2: A redox titration

    25.00 mL of a solution containing Fe²⁺ is acidified and titrated with 0.0200 M KMnO₄; the solution turns pale pink after 18.40 mL. Find the concentration of Fe²⁺.

    Step-by-step solution

    1. Write the mole ratio. In acid, MnOX4X−\ce{MnO4-} is reduced to MnX2+\ce{Mn^2+} (gaining 5 electrons) and FeX2+\ce{Fe^2+} is oxidized to FeX3+\ce{Fe^3+} (losing 1). Balancing electrons:

      MnOX4X−+5 FeX2++8 HX+→MnX2++5 FeX3++4 HX2O\begin{aligned} &\ce{MnO4- + 5Fe^2+ + 8H+} \\ &\ce{-> Mn^2+ + 5Fe^3+ + 4H2O} \end{aligned}

      MnOX4X−\ce{MnO4-} and FeX2+\ce{Fe^2+} react in a 1 : 5 mole ratio.

    2. Moles of MnOX4X−\ce{MnO4-}. In mol:

      n(MnOX4X−)=0.0200×0.01840=3.68×10−4\begin{aligned} n(\ce{MnO4-}) &= 0.0200 \times 0.01840 \\ &= 3.68 \times 10^{-4} \end{aligned}

    3. Convert to FeX2+\ce{Fe^2+} with the 1 : 5 ratio.

      n(FeX2+)=5×3.68×10−4=1.84×10−3\begin{aligned} n(\ce{Fe^2+}) &= 5 \times 3.68 \times 10^{-4} \\ &= 1.84 \times 10^{-3} \end{aligned}

    4. Concentration of FeX2+\ce{Fe^2+}.

      [FeX2+]=1.84×10−30.02500=0.0736 M\begin{aligned} [\ce{Fe^2+}] &= \frac{1.84 \times 10^{-3}}{0.02500} \\ &= \boxed{0.0736\ \text{M}} \end{aligned}

    5. The end point. MnOX4X−\ce{MnO4-} is purple and MnX2+\ce{Mn^2+} nearly colorless; once the FeX2+\ce{Fe^2+} is used up, one extra drop of MnOX4X−\ce{MnO4-} turns the solution pale pink. KMnOX4\ce{KMnO4} indicates its own end point, so no indicator is added.

    Practice: an unknown acid is titrated with standard NaOH. How does each problem affect the calculated acid concentration?
    1. The buret wasn't rinsed with the NaOH solution
    2. The flask was still wet with distilled water
    3. The flask was rinsed with the acid being analyzed
    4. An air bubble in the buret tip disappears during the titration
    5. The final buret reading is read from above the meniscus
    6. The initial buret reading is read from below the meniscus
    7. The end point is overshot, and the solution turns deep pink
    8. Some acid splashes out of the flask while swirling
    9. The NaOH was standardized with wet KHP
    10. A drop is left hanging on the buret tip at the end point

    Common mistakes

    Mistake 1: “The equivalence point is where the pH is 7.” That’s true only for a strong acid titrated with a strong base. A weak acid titrated with a strong base has an equivalence-point pH above 7; a weak base titrated with a strong acid, below 7.

    Mistake 2: “A weak acid only partly ionizes, so it never reaches the equivalence point.” OH⁻ reacts essentially completely with weak-acid molecules, so equal moles of a weak and a strong acid use the same amount of base.

    Mistake 3: “The end point is the equivalence point.” The end point is where the indicator changes color; the equivalence point is set by moles. Only with a suitable indicator are the two very close.

    Quick check

    25.00 mL of 0.100 M HCl and 25.00 mL of 0.100 M CH₃COOH are each titrated with 0.100 M NaOH. How do the volumes of NaOH needed to reach the equivalence point compare?

    Which titration technique is correct?

    Four titrations use 24.86 mL, 24.90 mL, 26.95 mL, and 24.88 mL of NaOH. How should the data be handled?

    Wet KHP is used to standardize a NaOH solution. How is the calculated NaOH molarity affected, and why?

    Acidified Fe²⁺ solution is titrated with KMnO₄ solution, with no indicator added. How is the end point recognized?

    How does the pH change during the titration of Fe²⁺ with KMnO₄ (MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O)?

    The Mg(OH)₂ in an antacid tablet is titrated with 0.500 M HCl, reaching the equivalence point at 24.60 mL: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O. What mass of Mg(OH)₂ is in the tablet?

    Which apparatus is best for measuring the volume of H₂ produced when magnesium reacts with hydrochloric acid?

    AP-style free response

    AP-style free response

    A student titrates 20.00 mL of hydrochloric acid of unknown concentration with 0.1000 M NaOH, using phenolphthalein as the indicator. The volumes of NaOH used in four trials are:

    Trial1234
    V(NaOH) (mL)21.4221.3823.1021.40

    (a) Explain the difference between the equivalence point and the end point. What is observed at the end point in this experiment?

    (b) Which trials should be used? Calculate the concentration of the hydrochloric acid.

    (c) If the buret had been washed with distilled water but not rinsed with the NaOH solution, would the calculated HCl concentration be too high or too low? Justify your answer.

    (d) If the analyte were 20.00 mL of acetic acid of the same concentration, what volume of NaOH would reach the equivalence point? Would the pH at the equivalence point be equal to, greater than, or less than 7?

    Model answer and scoring

    (a) The equivalence point is where the moles of OH⁻ added exactly equal the moles of H⁺ present; the end point is where the indicator changes color and the titration stops. Here, at the end point the solution turns from colorless to pale pink after the last drop of NaOH, and the color lasts for half a minute.

    (b) Trial 3 (23.10 mL) is clearly out of line and is discarded. Average volume = (21.42 + 21.38 + 21.40) ÷ 3 = 21.40 mL. [HCl] = 0.1000 × 21.40 ÷ 20.00 = 0.1070 M.

    (c) Too high. The leftover water dilutes the NaOH, so a larger volume is needed to reach the end point, and [HCl] = M(NaOH)V(NaOH) ÷ V(HCl) comes out too high.

    (d) The acetic acid has the same number of moles as the HCl, so it also needs 21.40 mL. At the equivalence point the solution contains CH₃COO⁻, a weak base, so the pH is greater than 7.

    Scoring: (a) 1 point for each definition and 1 for the end-point observation; (b) trial 3 discarded, 0.1070 M; (c) too high, explained by the larger volume; (d) 21.40 mL, pH greater than 7.

    Summary

    • The titrant is in the buret and the analyte in the flask; the equivalence point is set by moles, and the end point is shown by the indicator.
    • Equal moles of a monoprotic strong acid and weak acid use the same amount of base, but the pH at the equivalence point differs.
    • KMnO₄ titrates Fe²⁺ in a 1 : 5 ratio with no added indicator, turning pale pink at the end point; the reaction uses up H⁺, so the pH rises.
    • Rinse the buret with titrant; don’t rinse the flask; discard clearly out-of-line data; a wet primary standard makes the standardized molarity too high.

    AP tip: For titration-error questions, first decide which measured quantity (a volume or a mass) the mistake makes too large or too small, then follow it through the calculation to get the direction of the error. Naming the quantity that changed earns more credit than “too high” or “too low” alone.

    Practice this chapter: Unit 4 · Chemical Reactions · 12 questionsPractice this topic: U4.6 · Introduction to Titration · 7 questions

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