16.7

Titration Calculations: Weak Acid - Strong Base

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The pH at different stages of a weak acid or base titration is calculated using different methods during various points of the titration.

If a weak acid or base is the main determinant of the pH, the K_a or K_b and an ICE table, or the Henderson-Hasselbalch equation are used. If strong acid or base is present after the neutralization reaction, the concentration of the remaining hydronium or hydroxide ions is used to calculate the pH.

The initial pH of 50 mL of a 0.10 M acetic acid solution is 2.87 and is calculated using the K_a and an ICE table, as acetic acid is the main contributor.

If this solution containing 0.0050 moles of acetate is titrated with 0.10 M sodium hydroxide, the hydroxide ions react with acetic acid to produce acetate, resulting in a buffer.

Therefore, when 10 mL of the sodium hydroxide containing 0.0010 moles of hydroxide ions is added, 0.0010 moles of acetate are formed, and 0.0040 moles of acetic acid remain. The pH of the buffer can be calculated by substituting these values into the Henderson-Hasselbalch equation and equals 4.14.

When 25 mL of the sodium hydroxide is added, half of the initial moles of acetic acid are converted into acetate. At this point, the pH equals the pK_a, as the amount of acetic acid and acetate ion are equal.

The further addition of sodium hydroxide up to 50 mL converts all of the acetic acid molecules into acetate, and the equivalence point is reached. As acetate ions are basic, the equivalence point lies in the basic region.

The concentration of the acetate ion is calculated by dividing the number of moles by the total volume of the solution.

The pH is determined using the K_b for acetate ions and an ICE table, as the acetate ion is the main contributor to the pH at the equivalence point.

The K_b for acetate is calculated using the formula, K_w = K_a × K_b, and equals 5.6 × 10⁻¹⁰.

Substituting equilibrium concentrations in the expression for K_b gives the hydroxide concentration, 5.3 × 10⁻⁶ M. The pOH and pH of the solution are 5.28 and 8.72, respectively.

Further addition of sodium hydroxide in the solution results in a mixture of acetate ions and sodium hydroxide. However, the final concentration of sodium hydroxide determines the pH of the solution, as it is a stronger base than acetate.

Therefore, if 70 mL of sodium hydroxide is added into the solution, the final concentration of the hydroxide ions can be calculated by subtracting the total moles of acetic acid, 0.0050 moles, from the total moles of hydroxide ions added, 0.0070 moles, and dividing it by the total volume of the solution, 120 mL or 0.12 L.

As the hydroxide ion concentration is 0.017 M, the pOH and pH of the solution are calculated to be 1.78 and 12.22, respectively.

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Calculating pH for Titration Solutions: Weak Acid/Strong Base

For the titration of 25.00 mL of 0.100 M CH₃CO₂H with 0.100 M NaOH, the reaction can be represented as:

Chromatography diagram with solvent flow direction showing components' separation process.

The pH of the titration solution after the addition of the different volumes of NaOH titrant can be calculated as follows:

(a) The initial pH is computed for the acetic acid solution in the usual ICE approach:

Static equilibrium; ΣFx=0, ΣFy=0 equations; vector diagram; physics principles; balance analysis.

(b) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of

chemical reaction mechanism diagram showing transformation of reactants to products via intermediates

Base ionization of acetate is represented by the equation

Static equilibrium; equations, ΣFx=0, ΣFy=0; free-body diagram; force vector analysis.

Assuming x << 0.0500, the pH may be calculated via the usual ICE approach:

Ligand exchange process formulas; chemical reaction diagram.

Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.

(c) Titrant volume = 12.50 mL. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:

Peptide bond formation diagram showing amino acids linked by covalent bonds, illustrating protein synthesis.

(pH = pKa at the half-equivalence point in a titration of a weak acid)

(d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base:

NMR spectroscopy graph; chemical shifts vs. intensity; aromatic compound analysis results.

This text is adapted from Openstax, Chemistry 2e, Section 14.7: Acid-base Titrations.

Transcript

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The pH at different stages of a weak acid or base titration is calculated using different methods during various points of the titration.

If a weak acid or base is the main determinant of the pH, the K_a or K_b and an ICE table, or the Henderson-Hasselbalch equation are used. If strong acid or base is present after the neutralization reaction, the concentration of the remaining hydronium or hydroxide ions is used to calculate the pH.

The initial pH of 50 mL of a 0.10 M acetic acid solution is 2.87 and is calculated using the K_a and an ICE table, as acetic acid is the main contributor.

If this solution containing 0.0050 moles of acetate is titrated with 0.10 M sodium hydroxide, the hydroxide ions react with acetic acid to produce acetate, resulting in a buffer.

Therefore, when 10 mL of the sodium hydroxide containing 0.0010 moles of hydroxide ions is added, 0.0010 moles of acetate are formed, and 0.0040 moles of acetic acid remain. The pH of the buffer can be calculated by substituting these values into the Henderson-Hasselbalch equation and equals 4.14.

When 25 mL of the sodium hydroxide is added, half of the initial moles of acetic acid are converted into acetate. At this point, the pH equals the pK_a, as the amount of acetic acid and acetate ion are equal.

The further addition of sodium hydroxide up to 50 mL converts all of the acetic acid molecules into acetate, and the equivalence point is reached. As acetate ions are basic, the equivalence point lies in the basic region.

The concentration of the acetate ion is calculated by dividing the number of moles by the total volume of the solution.

The pH is determined using the K_b for acetate ions and an ICE table, as the acetate ion is the main contributor to the pH at the equivalence point.

The K_b for acetate is calculated using the formula, K_w = K_a × K_b, and equals 5.6 × 10⁻¹⁰.

Substituting equilibrium concentrations in the expression for K_b gives the hydroxide concentration, 5.3 × 10⁻⁶ M. The pOH and pH of the solution are 5.28 and 8.72, respectively.

Further addition of sodium hydroxide in the solution results in a mixture of acetate ions and sodium hydroxide. However, the final concentration of sodium hydroxide determines the pH of the solution, as it is a stronger base than acetate.

Therefore, if 70 mL of sodium hydroxide is added into the solution, the final concentration of the hydroxide ions can be calculated by subtracting the total moles of acetic acid, 0.0050 moles, from the total moles of hydroxide ions added, 0.0070 moles, and dividing it by the total volume of the solution, 120 mL or 0.12 L.

As the hydroxide ion concentration is 0.017 M, the pOH and pH of the solution are calculated to be 1.78 and 12.22, respectively.

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