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: pain and is a weak base. What is the
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 problems when simplifying assumptions are invalid
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.
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 NH-] and use this to calculate the pH
Example: A buffer made from 0.10 mol HA (p is within one unit of the 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: is within one unit of the p
Example: What is the pH and [CO is within one unit of the p32-] 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 very near the 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 very near the 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.