ACID BASE TITRATION

Acid-Base Titrations

An acid-base titration is when you add a base to an acid until the equivalence point is reached which is where the number of moles of acid equals the number of moles of base. For the titration of a strong base and a strong acid, this equivalence point is reached when the pH of the solution is seven (7) as seen on the following titration curve: For the titration of a strong base with a weak acid, the equivalence point is reached when the pH is greater than seven (7). The half equivalence point is when half of the total amount of base needed to neutralize the acid has been added. It is at this point where the pH = pK_a of the weak acid.
In an acid-base titration, the base will react with the weak acid and form a solution that contains the weak acid and its conjugate base until the acid is completely gone. To solve these types of problems, we will use the K_a value of the weak acid and the molarities in a similar way as we have before. Before demonstrating this way, let us first examine a short cut, called the Henderson-Hasselbalch Equation. This can only be used when you have some acid and some conjugate base in your solution. If you only have acid, then you must do a pure K_a problem and if you only have base (like when the titration is complete) then you must do a K_b problem.

pH = pKa + log

[base] [acid]

Where:
pH is the log of the molar concentration of the hydrogen
pK_a is the equilibrium dissociation constant for an acid
[base] is the molar concentration of a basic solution
[acid] is the molar concentration of an acidic solution

Example Problem: 25.0 mL of 0.400 M KOH is added to 100. mL of 0.150 M benzoic acid, HC₇H₅O₂ (K_a=6.3x10^-5). Determine the pH of the solution.

Answer:

Determine where in the titration we are:

0.400 M x 0.025 L = 0.0100 mol KOH added

0.150 M x 0.100 L = 0.0150 mol HC7H5O2 originally

because only 0.100 mol of base has been added, that means the thitration is not     complete; this means there are two ways to solve this problem: the normal way     and the way using the Henderson-Hasselbalch Equation.

Normal way:

HC7H5O2 + OH- C7H5O2- + H2O     before reaction: 0.015 mol 0.0100 mol 0 mol --     change: -0.0100 -0.0100 +0.0100 --     after reaction: 0.0050 0 0.0100 --

Ka = [H+][C7H5O2-] [HC7H5O2] = 6.3x10-5 = (x)(0.0100) 0.0050

x = [H+] = 3.2x10-5 M

pH = -log(3.2x10-5) = 4.49

Henderson-Hasselbalch Way:

[HC7H5O2] = 0.0050 mol 0.125 L = 0.040 M

[C7H5O2-] = 0.0100 mol 0.125 L = 0.0800 M

pH = pKa + log [base] [acid]

pH = -log(6.3x10-5) + log 0.0800 0.0400 = 4.20 + 0.30 = 4.50

This equation is used frequently when trying to find the pH of buffer solutions. A buffer solution is one that resists changes in pH upon the addition of small amounts of an acid or a base. They are made up of a conjugate acid-base pair such as HC₂H₃O₂/C₂H₃O₂^- or NH₄⁺/NH₃. They work because the acidic species neutralize the OH^- ions while the basic species neutralize the H⁺ ions. The buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to a significant degree. This depends on the amount of acid or base in the buffer. High buffering capacities come from solutions with high concentrations of the acid and the base and where these concentrations are similar in value. Practice weak acid problem:
C₆H₅COONa is a salt of a weak acid C₆H₅COOH. A 0.10 M solution of C₆H₅COONa has a pH of 8.60.

1. Calculate [OH^-] of C₆H₅COONa
2. Calculate K_b for: C₆H₅COO^- + H₂O <---> C₆H₅COOH + OH^-
3. Calculate K_a for C₆H₅COOH

See the weak acid solution.
Practice titration problem:
20.00 mL of 0.160 M HC₂H₃O₂ (K_a=1.8x10^-5) is titrated with 0.200 M NaOH.

1. What is the pH of the solution before the titration begins?
2. What is the pH after 8.00 mL of NaOH has been added?
3. What is the pH at the equivalence point?
4. What is the pH after 20.00 mL of NaOH has been added