Last Lecture:

Acid Equilibria

pH cont.

Note that the "p" has the more general meaning of "-log[]". Thus pOH is -log [OH^-], pCa = -log [Ca²⁺], etc.

pH of weak acid solutions

Weak acid dissociations involve equilibria. The equilibrium constants for weak acids have a specific symbol = K_a = K_eq for the dissociation of an acid.

With this in mind let's do some examples.

Example. Calculate the pH of a "buffer" (a solution which resists changes in pH) made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. K_a = 1.8 x 10^-5

Let's try the example again using the Henderson Hasselbalch equation, pH = pK_a + log[A^-] / [HA]

Example: Calculate the pH of a "buffer" made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. pK_a = 4.74. Notice that we are already past the "reaction" stage, so just set up for "@ Equilibrium." let's do the problem.

Notice that only the ratio matters, the actual concentrations of the acid and salt are not that important for determining pH!

Example. What ratio of acetate ion to acetic acid would you need to make up an acetate buffer with a pH of 5.25.? pK_a = 4.74. Easiest way to approach this is to use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc^-

Example: How many moles of sodium acetate must be added to 0.120 moles of acetic and sufficient water to make a liter if we want a buffer with a pH of 4.50? pK_a = 4.74.

Additional examples are available via the link below:

pH, Acid/Base, and Buffer Exercises

Colligative Properties

Colligative properties are those properties which depend only on the number or concentration, not on the type, of particles.

Be able to solve problems for:

• freezing point depression: DT_f = -k_fm, where m = molality = moles solute/kg solvent, and k_f is a constant specific to the solvent. (Recall MW lab as example)

   

Colligative Properties Exercises