Volumetric Methods and Neutralization Titrations

Volumetric Methods

• Titration methods rely on determining the quantity of a reagent with a known concentration to react completely with the analyte.
• The reagent can be a standard chemical solution or an electric current of known magnitude.
• Volumetric titrations measure the volume of a standard reagent.
• Gravimetric titrations measure the mass of the reagent instead of its volume.
• Coulometric titrations measure the quantity of charge required to react with the analyte.
• Redox titrations use volumetric methods involving electron transfer in analytical reactions.
• Additional methods include amperometric and spectrophotometric titrations.

Terms Used in Volumetric Titrations

• A standard solution (or standard titrant) is a reagent with a precisely known concentration.
  • In titration, a standard solution is added slowly from a buret to the analyte until the reaction is complete.
  • The required reagent volume or mass is the difference between initial and final readings.
• Back-titration determines the excess of a standard solution used to consume an analyte by titrating with a second standard solution.
  • Useful when the reaction rate is slow or the standard solution is unstable.

Equivalence Points and End Points

• The equivalence point is when the amount of added standard reagent is stoichiometrically equal to the amount of analyte.
• The end point is the point in a titration where a physical change indicates chemical equivalence.
• Indicators produce a physical change near the equivalence point.

Primary Standards

• A primary standard is a highly purified reference material with the following characteristics:
  1. High purity
  2. Atmospheric stability
  3. Absence of hydrate water
  4. Modest cost
  5. Reasonable solubility in the titration medium
  6. Reasonably large molar mass
• Secondary standards are less pure and their purity is determined by chemical analysis.

Standard Solutions

• The ideal standard solution should:
  1. Be sufficiently stable
  2. React rapidly with the analyte
  3. React completely with the analyte
  4. Undergo a selective reaction with the analyte, described by a balanced equation
• Standard solution concentrations are established using two methods:
  • Direct method: A carefully measured mass of a primary standard is dissolved in a suitable solvent and diluted to a known volume.
  • Standardization: The concentration of a volumetric solution is determined by titrating it against a carefully measured quantity of a primary or secondary standard, or an exactly known volume of another standard solution.
• A titrant that is standardized is sometimes called a secondary-standard solution.
• The direct method is preferred due to the larger uncertainty in secondary-standard solutions.

Volumetric Calculations

• Concentrations of standard solutions are typically expressed as molar concentration ($c$) or normal concentration.
  • Molar concentration is the number of moles of reagent in 1 liter of solution.
  • Normal concentration is the number of equivalents of reagent in 1 liter of solution.

Useful Relationships

• Most volumetric calculations rely on two pairs of equations:
  • Equation 11-1: $C_W(A) = \frac{\text{amount A (mol)}}{\text{no. kg solution}} = \frac{\text{amount A (mmol)}}{\text{no. g solution}}$
  • Equation 11-2: $C_W(A) = \frac{\text{mass A (g)}}{\text{molar mass A (g/mol)}}$
  • Equation 11-3: $\text{no. mol A} = \frac{\text{no. mol A}}{n_A}$
  • Equation 11-4: $\text{amount A (mmol)} = V(mL) \times C_A = \frac{\text{mmol A}}{\text{mL}}$

Calculating Molar Concentration of Standard Solutions

• The number of moles of a compound is calculated by multiplying the volume of the solution in liters by the concentration of the compound in moles per liter.
• Example 11-1: Preparing 2.000 L of 0.0500 M $K<em>2Cr</em>2O<em>7$ (169.87 g/mol). $\text{mass } K</em>2Cr<em>2O</em>7 = 2.000 \text{ L} \times 0.0500 \frac{\text{mol}}{\text{L}} \times 169.87 \frac{\text{g}}{\text{mol}} = 16.987 \text{ g}$
• Dissolve 16.987 g of $K<em>2Cr</em>2O_7$ in water and dilute to 2.000 L in a volumetric flask.
• Example 11-2: Preparing 500 mL of 0.0100 M $Na^+$ solution from $Na<em>2CO</em>3$ (105.99 g/mol).
• Since $Na<em>2CO</em>3$ dissociates into two $Na^+$ ions:
  $\text{mmol } Na<em>2CO</em>3 = 500 \text{ mL} \times 0.0100 \frac{\text{mmol } Na^+}{\text{mL}} \times \frac{1 \text{ mmol } Na<em>2CO</em>3}{2 \text{ mmol } Na^+} = 2.50 \text{ mmol } Na<em>2CO</em>3$

$\text{mass } Na<em>2CO</em>3 = 2.50 \text{ mmol} \times 105.99 \frac{\text{mg}}{\text{mmol}} = 265 \text{ mg} = 0.265 \text{ g}$

• Dissolve 0.265 g of $Na<em>2CO</em>3$ in water and dilute to 500 mL.

Working with Titration Data

• Calculations based on Equations 11-2 and 11-4, and the stoichiometric ratio between analyte and titrant.
• Example 11-4: A 50.00-mL portion of HCl solution requires 29.71 mL of 0.01963 M $Ba(OH)_2$ to reach an endpoint.
• 1 mmol $Ba(OH)_2$ reacts with 2 mmol HCl.
• Stoichiometric ratio: $\frac{2 \text{ mmol HCl}}{1 \text{ mmol } Ba(OH)<em>2}$$\text{mmol } Ba(OH)</em>2 = 29.71 \text{ mL} \times 0.01963 \frac{\text{mmol } Ba(OH)<em>2}{\text{mL}} = 0.5832 \text{ mmol } Ba(OH)</em>2$
  $\text{mmol HCl} = 0.5832 \text{ mmol } Ba(OH)<em>2 \times \frac{2 \text{ mmol HCl}}{1 \text{ mmol } Ba(OH)</em>2} = 1.166 \text{ mmol HCl}$
• Molar concentration of HCl:
  $\text{Molarity HCl} = \frac{1.166 \text{ mmol}}{50.00 \text{ mL}} = 0.0233 \text{ M}$
• Example 11-6: A 0.8040-g sample of iron ore is dissolved in acid, reduced to $Fe^{2+}$, and titrated with 47.22 mL of 0.02242 M $KMnO_4$ solution.
• Reaction: $MnO<em>4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H</em>2O$
• Molar mass of Fe is 55.847 g/mol, and $Fe<em>3O</em>4$ is 231.54 g/mol.
• % Fe calculation:
  $\text{amount } KMnO<em>4 = 47.22 \text{ mL } KMnO</em>4 \times 0.02242 \frac{\text{mmol } KMnO<em>4}{\text{mL } KMnO</em>4} = 1.058 \text{ mmol } KMnO<em>4$$\text{amount Fe} = 1.058 \text{ mmol } KMnO</em>4 \times \frac{5 \text{ mmol Fe}}{1 \text{ mmol } KMnO_4} = 5.290 \text{ mmol Fe}$
  $\text{mass Fe} = 5.290 \text{ mmol Fe} \times 55.847 \frac{\text{g}}{\text{mol}} = 0.2955 \text{ g Fe}$
  \text{% Fe} = \frac{0.2955 \text{ g Fe}}{0.8040 \text{ g sample}} \times 100% = 36.76%
• % $Fe<em>3O</em>4$ calculation:
  $\text{stoichiometric ratio} = \frac{5 \text{ mmol } Fe<em>3O</em>4}{3 \text{ mmol } KMnO<em>4}$$\text{amount } Fe</em>3O<em>4 = (47.22 \times 0.02242) \text{ mmol } KMnO</em>4 \times \frac{5 \text{ mmol } Fe<em>3O</em>4}{3 \text{ mmol } KMnO<em>4}$$\text{mass } Fe</em>3O<em>4 = 47.22 \times 0.02242 \times \frac{5}{3} \times 0.23154 \text{ g } Fe</em>3O<em>4$ \text{% } Fe3O4 = \frac{47.22 \times 0.02242 \times \frac{5}{3} \times 0.23154 \text{ g } Fe3O_4}{0.8040 \text{ g sample}} \times 100% = 50.81%

Titration Curves

• End points are detected by:
  • Color changes due to the reagent, analyte, or an indicator
  • Changes in potential of an electrode that responds to the titrant or analyte concentration.
• Titration curves are plots of a concentration-related variable versus titrant volume.

Types of Titration Curves

• Sigmoidal curve: p-function of analyte (or titrant) is plotted against titrant volume.
• Linear segment curve: Measurements are made on both sides, away from the equivalence point. The vertical axis represents instrument reading proportional to the concentration of the analyte or the titrant.
• Sigmoidal curves offer speed and convenience.
• Linear segment curves are useful for reactions that are only complete with a considerable excess of reagent or analyte.

Concentration Changes During Titrations

• Table 11-1 illustrates the changes in the relative concentrations of reagent and analyte at the equivalence point in a titration.

$[H<em>3O^+] = [HCl]$$[HCl]=\frac{\text{original mmol HCl - mmol NaOH added}}{\text{total volume of the solution}}$$[H</em>3O^+] = \frac{(50.00 \text{ mL} \times 0.1000 \text{ M}) - (V<em>{NaOH} \times 0.1000 \text{ M})}{50.00 \text{ mL} + V</em>{NaOH}}$
$V<em>{NaOH} = \frac{5.00 - [H</em>3O^+]50.00}{0.1000 + [H_3O^+]}$

• Beyond the equivalence point: $\text{pOH} = -\log[OH^-]$

Principles of Neutralization Titrations

Standard Solutions

• Standard solutions in neutralization titrations use strong acids or strong bases, which react more completely with the analyte and provide sharper end points.
• Standard acid solutions are prepared by diluting concentrated hydrochloric, perchloric, or sulfuric acid.
  • Nitric acid is rarely used.
• Standard base solutions are usually prepared from solid sodium, potassium, and sometimes barium hydroxides.

Acid-Base Indicators

• An acid-base indicator is a weak organic acid or base where the undissociated form has a different color than its conjugate form.
• $HIn + H<em>2O \rightleftharpoons H</em>3O^+ + In^-$
• $K<em>a = \frac{[H</em>3O^+][In^-]}{[HIn]}$
  $\frac{[HIn]}{[In^-]} = \frac{[H<em>3O^+]}{K</em>a}$
• Equation 12-3: $\text{pH} = \text{p}K_a \pm 1$
• Determinate error occurs when the indicator's color change differs from the equivalence point pH. Minimize by selecting the indicator carefully or by making a blank correction.
• Indeterminate error results from the human eye's limited ability to distinguish the indicator's intermediate color.

Variables that Influence Indicator Behavior:

• Temperature
• Ionic strength of the medium
• Presence of organic solvents and colloidal particles

Titration of Strong Acids and Bases

• Hydronium ions in a strong acid solution come from the acid's reaction with water and the dissociation of water itself.

Titrating a Strong Acid with a Strong Base

• Distinguish between calculated hypothetical titration curves and experimental curves.
• Three types of calculations are needed to construct the hypothetical curve:
  1. Preequivalence: Compute the acid concentration from its initial concentration and the amount of added base.
  2. Equivalence point: Hydronium and hydroxide ions are present in equal concentrations and the hydronium ion concentration can be calculated directly from the ion-product constant for water, $K_w$.
  3. Postequivalence: Compute the analytical concentration of the excess base, and the hydroxide ion concentration is approximately equal to the analytical concentration.
• Calculating the volume of NaOH

$[H<em>3O^+] = [HCl]$$[HCl]=\frac{\text{original mmol HCl - mmol NaOH added}}{\text{total volume of the solution}}$$[H</em>3O^+] = \frac{(50.00 \text{ mL} \times 0.1000 \text{ M}) - (V<em>{NaOH} \times 0.1000 \text{ M})}{50.00 \text{ mL} + V</em>{NaOH}}$
$V<em>{NaOH} = \frac{5.00 - [H</em>3O^+]50.00}{0.1000 + [H_3O^+]}$

• Challenge use the same reasoning to show that beyond the equivalence point

Titrating a Strong Base with a Strong Acid

*Curves for the titration of 0.0500 M and 0.00500 M NaOH with 0.1000 M and 0.0100 M HCl, respectively, are shown in Figure 12-5.

• Use the same criteria described for the titration of a strong acid with a strong base to select an indicator.

Titration Curves for Weak Acids

• Four types of calculations are needed to compute values for a weak acid or weak base titration curve.

  1. Initially, only a weak acid or weak base is present, and the pH is calculated from the concentration of that solute and its dissociation constant.
  2. After various increments of titrant have been added (up to, but not including, the equivalence point), the solution consists of a series of buffers. The pH of each buffer can be calculated from the analytical concentrations of the conjugate base or acid and the concentrations of the weak acid or base that remains.
  3. At the equivalence point, the solution contains only the conjugate of the weak acid or base being titrated and the pH is calculated from the concentration of this product.
  4. Beyond the equivalence point, the excess of strong acid or base titrant suppresses the acidic of basic character of the reaction product to such an extent that the pH is governed largely by the concentration of the excess titrant.

• Example 12-3: Titration of 50.00 mL of 0.1000 M acetic acid (HOAc) with 0.1000 M sodium hydroxide at 25°C.

  • Initial pH:
    $pH = \frac{pK_a - \log[HA]}{2}$

• After Addition of 10.00 mL of Reagent
  $[NaOAc] = \frac{10.00 \text{ mL } \times 0.1000 \text{ M}}{50.00 \text{ mL} + 10.00 \text{ mL}}$
  $[HOAc] = \frac{50.00 \text{ mL } \times 0.1000 \text{ M} - 10.00 \text{ mL } \times 0.1000 \text{ M}}{50.00 \text{ mL} + 10.00 \text{ mL}}$
  $K<em>a = \frac{[H</em>3O^+][OAc^-]}{[HOAc]}$

• After Addition of 25.00 mL of Reagent
  $[NaOAc] = \frac{25.00 \text{ mL } \times 0.1000 \text{ M}}{50.00 \text{ mL} + 25.00 \text{ mL}}$
  $[HOAc] = \frac{50.00 \text{ mL } \times 0.1000 \text{ M} - 25.00 \text{ mL } \times 0.1000 \text{ M}}{50.00 \text{ mL} + 25.00 \text{ mL}}$
  $K<em>a = \frac{[H</em>3O^+][OAc^-]}{[HOAc]}$

• Equivalence-Point pH
  $[NaOAc] = \frac{50.00 \text{ mL } \times 0.1000 \text{ M}}{50.00 \text{ mL} + 50.00 \text{ mL}} = 0.0500 \text{ M}$

• After Addition of 50.10 mL of Base
  [OH-] = (0.10 \text{ mL } \times 0.1000 \text{ M})/(100.10 \text{ mL} ) = 1.00 \times 10^{-4} \text{ M}

• At the half-titration point in a weak-acid titration, the hydronium ion concentration equals $K_a$.

• At the half-titration point in a weak-base titration, the hydroxide ion concentration equals $K_b$.

• The buffer capacities of each of the solutions are at a maximum at this point.

The Effect of Concentration

• The pH of buffers is largely independent of dilution.

The Effect of Reaction Completeness

• The pH change in the equivalence-point region becomes smaller as the acid becomes weaker and the reaction becomes less complete.

Choosing an Indicator: The Feasibility of Titration

• The choice of indicator is more limited for the titration of a weak acid than the titration of a strong acid.

Titration Curves for Weak Bases

• Indicators with mostly acidic transition ranges must be used for weak bases.

The Composition of Solutions During Acid Base Titrations