Acid-Base Titrations酸碱滴定与滴定曲线

Big Idea · Structure and Properties (SAP)AP Science Practice 4 · Model AnalysisAP link · Unit 8~50 min

Learning goals

  • Read a titration curve (starting point, buffer region, half-equivalence point, equivalence point and excess region), and explain what sets the pH in each part
  • Find the concentration of the analyte from the volume at the equivalence point, and find Ka from the pH at the half-equivalence point (pH = pKa)
  • Compare titration curves for strong acids, weak acids, weak bases and polyprotic acids, and choose a suitable indicator

Before you start

  • Neutralization and titration (2.12.5)
  • Introduction to titration (U4.6)
  • Acid-base reactions and buffers (U8.4)

Titration curves

In a titration, a solution of known concentration (the titrant) is added drop by drop to the solution being analyzed, and a plot of pH against the volume added gives the titration curve. The point where the added acid and base have reacted in exactly the ratio of the equation is the equivalence point化学计量点; near it the pH changes fastest, and the curve shows a steep “jump.”

When a strong base titrates a weak acid HA, the curve has four parts:

  1. Starting point: only the weak acid is present, so Ka sets the pH (U8.3), which is higher than for a strong acid of the same concentration.
  2. Buffer region: the added OH⁻ turns part of the HA into A⁻, so HA and A⁻ are both present. This is a buffer, and the pH changes slowly. When exactly half of the base needed for equivalence has been added, [HA] = [A⁻]; this is the half-equivalence point半中和点:
Ka=[HX3OX+][AX−][HA]=[HX3OX+]⟹pH=pKa\begin{gathered} K_a = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} = [\ce{H3O+}] \\ \Longrightarrow\quad \text{pH} = \text{p}K_a \end{gathered}
  1. Equivalence point: all the HA has become A⁻, which reacts with water to form OH⁻, so the pH is above 7.
  2. Excess region: the excess OH⁻ sets the pH, whatever the weak acid is.

When a strong acid titrates a weak base, it’s the other way round: at the equivalence point only the conjugate acid BH⁺ remains, so the pH is below 7; at the half-equivalence point the pH equals the pKa of BH⁺. A polyprotic acid (such as H₃PO₄) has one equivalence point for each proton it gives up, so its curve can show several jumps.

Click the key points below to see what the solution mainly contains at each point and what sets the pH:

Try it: key points on a titration curve

Choosing an indicator

An indicator is itself a weak acid whose acid and base forms have different colors, and it changes color over a range of roughly pKa ± 1. The point where the indicator changes color is the end point终点. Choose an indicator whose color-change range lies within the jump and includes or is close to the pH at the equivalence point, so that the end point nearly matches the equivalence point.

IndicatorColor-change range (pH)
Methyl orange3.1–4.4
Methyl red4.4–6.2
Bromothymol blue6.0–7.6
Phenolphthalein8.3–10.0

Compare the titration curves of the strong acid HCl and the weak acid CH₃COOH, and try different indicators:

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

    Example 1: concentration and Ka from titration data

    20.00 mL of a monoprotic weak acid HX is titrated with 0.120 M NaOH. The equivalence point is reached after 16.40 mL of NaOH; after 8.20 mL of NaOH, the pH is 3.75. Find the concentration of HX and its Ka.

    Step-by-step solution

    1. At the equivalence point, moles of OH⁻ equal moles of HX. HX + OH⁻ → X⁻ + H₂O reacts 1 : 1:

      n(HX)=n(OHX−)=0.01640×0.120=1.968×10−3 mol\begin{aligned} n(\ce{HX}) &= n(\ce{OH-}) \\ &= 0.01640 \times 0.120 \\ &= 1.968 \times 10^{-3}\ \text{mol} \end{aligned}

    2. Find the concentration of HX. Divide by the acid’s volume, 20.00 mL = 0.02000 L:

      [HX]=1.968×10−30.02000=0.0984 M\begin{aligned} [\ce{HX}] &= \frac{1.968 \times 10^{-3}}{0.02000} \\ &= \boxed{0.0984\ \text{M}} \end{aligned}

    3. What’s in the flask after 8.20 mL. 8.20 mL is exactly half of 16.40 mL, so the OH⁻ added has neutralized half the HX (in mmol):

      HXOH⁻X⁻
      Before1.9680.9840
      Change−0.984−0.984+0.984
      After0.98400.984

      The HX left equals the X⁻ formed: this is the half-equivalence point.

    4. At half-equivalence, pH = pKa. In the Ka expression, [X⁻] and [HX] cancel when they are equal:

      Ka=[HX3OX+] [XX−][HX]=[HX3OX+]\begin{aligned} K_a &= \frac{[\ce{H3O+}]\,[\ce{X-}]}{[\ce{HX}]} \\ &= [\ce{H3O+}] \end{aligned}

      Taking the negative log of both sides, pKa = pH = 3.75.

    5. Find Ka.

      Ka=10−3.75=1.8×10−4K_a = 10^{-3.75} = \boxed{1.8 \times 10^{-4}}

    Example 2: titrating NH₃ with HCl

    20.0 mL of 0.150 M NH₃ (Kb = 1.8 × 10⁻⁵) is titrated with 0.100 M HCl. Find the volume of HCl at the equivalence point and the pH there, and choose an indicator from the table above.

    Step-by-step solution

    1. Moles of NH₃.

      n(NHX3)=0.0200×0.150=3.00×10−3 mol\begin{aligned} n(\ce{NH3}) &= 0.0200 \times 0.150 \\ &= 3.00 \times 10^{-3}\ \text{mol} \end{aligned}

    2. Volume of HCl at the equivalence point. NH₃ + H₃O⁺ → NH₄⁺ + H₂O reacts 1 : 1:

      V(HCl)=3.00×10−30.100=0.0300 L=30.0 mL\begin{aligned} V(\ce{HCl}) &= \frac{3.00 \times 10^{-3}}{0.100} = 0.0300\ \text{L} \\ &= \boxed{30.0\ \text{mL}} \end{aligned}

    3. What’s in the flask at the equivalence point. All the NH₃ has become NH₄⁺ (3.00 × 10⁻³ mol), in a total of 20.0 + 30.0 = 50.0 mL (concentration in M):

      [NHX4X+]=3.00×10−30.0500=0.0600[\ce{NH4+}] = \frac{3.00 \times 10^{-3}}{0.0500} = 0.0600

    4. Ka of NH₄⁺.

      Ka=KwKb=1.0×10−141.8×10−5=5.6×10−10\begin{aligned} K_a &= \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \\ &= 5.6 \times 10^{-10} \end{aligned}

    5. Find [H₃O⁺] (in M). Set up an ICE table as for any weak acid; Ka is tiny, so x ≈ √(Ka·c):

      x≈5.6×10−10×0.0600=3.36×10−11=5.8×10−6\begin{aligned} x &\approx \sqrt{5.6 \times 10^{-10} \times 0.0600} \\ &= \sqrt{3.36 \times 10^{-11}} = 5.8 \times 10^{-6} \end{aligned}

    6. Find the pH.

      pH=−log⁡(5.8×10−6)=5.24\text{pH} = -\log(5.8 \times 10^{-6}) = \boxed{5.24}

    7. Choose an indicator. The indicator’s range should include the pH at the equivalence point. Methyl red’s range, 4.4–6.2, includes 5.24, so it works; phenolphthalein changes only above pH 8.3, so it isn’t suitable.

    Practice: from the pH at the equivalence point, choose the most suitable indicator.
    1. CH₃COOH titrated with NaOH, equivalence pH 8.72
    2. NH₃ titrated with HCl, equivalence pH 5.28
    3. HCl titrated with NaOH, equivalence pH 7.00
    4. HN₃ titrated with NaOH, equivalence pH 8.65
    5. A weak base titrated with HCl, equivalence pH 3.6
    6. H₃PO₄ titrated with NaOH to the first equivalence point, pH 4.70
    7. H₃PO₄ titrated with NaOH to the second equivalence point, pH 9.67
    8. (CH₃)₃N titrated with HCl, equivalence pH 5.55

    Common mistakes

    Mistake 1: “The pH at the equivalence point is always 7.” That’s true only for a strong acid titrated with a strong base. For a weak acid titrated with a strong base the equivalence pH is above 7; for a weak base titrated with a strong acid it is below 7.

    Mistake 2: “The weaker the acid, the less NaOH the titration needs.” The volume at the equivalence point depends only on the amount of acid. The same volume of HCl and CH₃COOH at the same concentration needs the same amount of NaOH; what differs is the pH at the start, in the buffer region and at the equivalence point.

    Mistake 3: “At the half-equivalence point half the weak acid has ionized, so [H₃O⁺] = [HA] ÷ 2.” At the half-equivalence point half the HA has been neutralized by OH⁻ to A⁻, so [HA] = [A⁻], which gives pH = pKa.

    Quick check

    A monoprotic weak acid is titrated with NaOH, and the pH at the half-equivalence point is 4.20. What is Ka for the acid?

    When NH₃ solution is titrated with HCl, the pH at the equivalence point is:

    Which indicator is least suitable for titrating CH₃COOH with NaOH?

    15.0 mL of 0.200 M oxalic acid, H₂C₂O₄ (a diprotic acid), is titrated with 0.150 M NaOH to the second equivalence point. What volume of NaOH is needed?

    25.0 mL of 0.10 M HCl and 25.0 mL of 0.10 M CH₃COOH are each titrated with 0.10 M NaOH. What is the same for both?

    When NH₃ is titrated with HCl, which nitrogen-containing species are mainly present in the buffer region?

    AP-style free response

    AP-style free response

    25.00 mL of 0.100 M hydrazoic acid, HN₃ (Ka = 2.5 × 10⁻⁵, OpenStax), is titrated with 0.100 M NaOH.

    (a) Calculate the pH of the HN₃ solution before any NaOH is added.

    (b) What volume of NaOH is needed to reach the equivalence point? What is the pH at the half-equivalence point?

    (c) Calculate the pH at the equivalence point.

    (d) Choose an indicator, methyl red (4.4–6.2) or phenolphthalein (8.3–10.0), and justify your choice.

    (e) Calculate the pH after 30.00 mL of NaOH has been added.

    Model answer and scoring

    (a) x ≈ √(2.5 × 10⁻⁵ × 0.100) = 1.6 × 10⁻³ M, so pH = 2.80 (x ÷ c ≈ 1.6%, so the approximation holds).

    (b) n(HN₃) = 2.50 mmol, so 25.00 mL of NaOH is needed. At the half-equivalence point (12.50 mL), [HN₃] = [N₃⁻] and pH = pKa = −log(2.5 × 10⁻⁵) = 4.60.

    (c) At the equivalence point only 2.50 mmol N₃⁻ remains in a total of 50.00 mL, so [N₃⁻] = 0.0500 M. Kb = Kw ÷ Ka = 4.0 × 10⁻¹⁰, so [OH⁻] ≈ √(4.0 × 10⁻¹⁰ × 0.0500) = 4.5 × 10⁻⁶ M, pOH = 5.35 and pH = 8.65.

    (d) Phenolphthalein. The equivalence pH of 8.65 lies within its range; methyl red changes color below pH 6.2, while the solution is still in the buffer region, so the end point would come too early.

    (e) OH⁻ is in excess by 0.100 × 5.00 = 0.500 mmol in a total of 55.00 mL, so [OH⁻] = 9.09 × 10⁻³ M, pOH = 2.04 and pH = 11.96.

    Scoring: (a) pH of the weak acid; (b) volume, and pH = pKa; (c) concentration of the conjugate base, Kb, and pH; (d) equivalence pH within the indicator’s range; (e) excess OH⁻ and total volume.

    Summary

    • Weak acid titrated with strong base: Ka sets the starting pH; HA and A⁻ coexist in the buffer region; pH = pKa at the half-equivalence point; pH > 7 at the equivalence point; after that, the excess OH⁻ sets the pH.
    • The volume at the equivalence point depends only on the amount of analyte; the pH there depends on the conjugate acid or base that remains.
    • An indicator’s color-change range should lie within the jump and include or be close to the equivalence pH.

    AP tip: To read a titration curve, find the equivalence volume first, then take half of it to find the half-equivalence point and read off the pKa. To explain the equivalence pH, write “only the conjugate base A⁻ remains: A⁻ + H₂O ⇌ HA + OH⁻.” When choosing an indicator, state that “the equivalence pH is within its color-change range.”

    Practice this chapter: Unit 8 · Acids and Bases · 12 questionsPractice this topic: U8.5 · Acid-Base Titrations · 8 questions

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