Chem 107 

Fall 2009 
Lecture Notes: pH and Acid/Base Examples 





Example. What is the pH of a 0.10 M solution of acetic acid. K_{a} = 1.8 x 10^{5}

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}

So now let's do the problem above using the HendersonHasselbalch equation, pH = pK_{a} + log[A^{}] / [HA].
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. pK_{a} = 4.74.. Notice that we are already past the "reaction" stage, so just set up for "@ Equilibrium."
HOAc H^{+} + OAc^{} @ Equilibrium
(0.0125 x) M assume x is small, = 0.0125 x
(0.0250  x) M assume x is small, = 0.0250 pH = pK_{a} + log[A^{}] / [HA] pH = 4.74 + log(0.250) / (0.125) = 4.74 + 0.301 pH = 5.04
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 HendersonHasselbalch equation to find the ratio of HOAc to OAc^{} pH = pK_{a} + log[A^{}] / [HA] Rearranging, log[A^{}] / [HA] = pH  pK_{a}
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.
Again use the HendersonHasselbalch equation to find the ratio of HOAc to OAc^{} pH = pK_{a} + log[A^{}] / [HA] Rearranging and substituting, log[OAc^{}] / [HOAc] = pH  pK_{a} Substituting, log[OAc^{}] / [HOAc] = 4.50  4.74 =  0.24 Raise both sides to power of 10: [OAc^{}] / [HOAc] = 0.575 [OAc^{}] =0.575 [HOAc]

© R A Paselk
Last modified 9 December 2005