Chapter 18: Equilibria in Solutions of Weak Acids and Bases. All weak acids behave the same way in aqueous solution: they partially ionize In terms of the “general” weak acid HA, this can be written as: Following the procedures in Chapter 16. K a is called the acid ionization constant
Relative strengths of conjugate acid-base pairs. The stronger the acid is, the weaker the conjugate base. The weaker the acid, the stronger the conjugate base. Very strong acids ionize 100% and their conjugate bases do not react to any measurable extent.
ANALYSIS: The reaction can be represented as:
At equilibrium, [OH-] = x = 10-pOH
SOLUTION: Use pOH = 14.00 – pH, substituting:
Determining how to proceed in acid-base equilibrium problems. The nature of the solute species determines how the problem is approached. This flowchart can help get you started in the right direction.
ANALYSIS: 400* Kb>0.0010 M, so use of the quadratic equation is indicated.
SOLUTION: Set the problem up
Put in standard form
Solve for x and the equilibrium concentrations
The method of successive approximations. Following these steps leads to a solution of equilibrium problems when the usual simplifying assumptions fail.
Using a subscript to keep track of the iterations the method gives:
Comment: This method works best when the initial guess is “close” to the final answer. At equilibrium, the base is 61% ionized, so starting with a guess of 0% ionized is “far” from the final solution.
Example: What is the pH of a buffer made by adding 0.10 mol NH3 and 0.11 mol NH4Cl to 2.0 L of solution? The Kb for ammonia is 1.8x10-5
ANALYSIS: This is a buffer, initial concentrations can be used as equilibrium values:
SOLUTION: Solve for [OH-] and use this to calculate the pH
Example: A buffer made from 0.10 mol HA (pKa=7.20)and 0.15 mol NaA in 2.0 L has 0.02 mol of HCl added to it with no volume change. What is the pH change?
ANALYSIS: This buffer problem is “best” solved in terms of moles. HCl is a strong acid. The H+ it contributes to solution increases the amount of HA present at the expense of A-.
SOLUTION: The pH before addition of HCl was:
After the HCl ionizes and reacts:
Example: What is the pH and [CO32-] in 0.10 M carbonic acid (H2CO3)?
ANALYSIS: Carbonic acid is a diprotic acid. The pH will depend on the [H+] generated from the first ionization. Ionization constants can be obtained from Table 18.3
SOLUTION: Solve the first ionization first, then substitute the results into the second reaction.
Titration curve for the titration of 25.00 mL of 0.2000 M HCl (a strong acid) with the 0.2000 M NaOH (a strong base). The equivalence point occurs at 25.00 mL added base with a pH of 7.0 (data from Table 18.4).
The titration curve for the titration of 25.00 mL of 0.200 M acetic acid with 0.200 M sodium hydroxide. Due to hydrolysis, the pH at the equivalence point higher than 7.00 (data from Table 18.5).
Titration curve for the titration of 25.00 mL of 0.200 M NH3 with 0.200 M HCl. The pH at the equivalence point is below 7.00 because of the hydrolysis of NH4+.
The titration of the diprotic acid H2A by a strong base. As each equivalence point is reached, the pH rises sharply.