Practice Problems 6.20.2011-KEY
1. Sodium hypochlorite (NaOCl, the active ingredient of all bleaches) was dissolved
in a solution buffered to pH 6.20. Find the ratio of [OCl-]/[HOCl] in this solution.
pKa for hypochloric acid is 7.53.
Since the pH of the solution is known, and the pKa of the solution is known, we
can easily calculate the [OCl-]/[HOCl] ratio in the solution using the HendersonHasselbalch equation:
base
pH  pK a  log
 acid 
OCl  
6.20  7.53  log
 HOCl 
OCl  
 0.047
 HOCl 
2. Find the pH of a solution prepared by dissolving 12.43 g of tris (MW = 121.135
g/mol) with 4.67 g of tris hydrochloride (MW = 157.596 g/mol) in 1.00 L of water.
pKa for tris hydrochloride is 8.075.
In this case, tris is the base, and tris hydrochloride is the acid. First, find the
concentrations of each in solution:
12.43g
[tris] 
 0.1026mol  0.1026M
121.135 g / mol
4.67 g
[tris  HCl ] 
 0.0296mol  0.0296M
157.596 g / mol
Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the
buffer solution:
base
pH  pK a  log
 acid 
pH  8.075  log
0.1026M
 8.61
0.0296M
3. If 12.0 mL of 1.00 M HCl is added to the solution in problem #2, what will the new
pH of the solution be?
To find the pH, we need to set up an ice table to determine the concentrations of
tris (base) and tris-HCl (acid) present at equilibrium. Recall that when a strong
acid is added to a buffer solution, all of the strong acid will react with the
conjugate base to form more of the weak acid. Another thing to note – the pH of
a buffer solution is independent of the volume – it will cancel out in the log
term… so we can do the entire calculation by just using the number of moles of
everything present!
For the reaction:
tris
+
H+
→
Tris-H+
Initial moles
0.1026
0.0120
0.0296
Change in moles
-0.0120
-0.0120
+0.0120
Final moles
0.0906
0
0.0416
Now the H.H. equation can be used once again to calculate the pH:
base
pH  pK a  log
 acid 
0.0906moles
 8.41
0.0416moles
Note that this is a small change in pH!
pH  8.075  log
*As another thing to think about, what would happen if that amount of strong
acid were added to just a solution of either tris or tris-HCl? What would happen
if a strong base were added to any of the solutions?