Untitled - Kelly Walsh High School

Buffers and Other Equilibria 245
Quick Tip
After the reaction, the moles of sodium hydroxide, limiting reagent, are zero;
therefore, sodium hydroxide no longer affects the pH. The solution only contains
unreacted nitrous acid, sodium ions, nitrite ions, and water. Cations coming
from a strong base do not affect the pH, so we do not need to worry about
the sodium ions. Water will not affect the pH either. The nitrous acid and its conjugate
base, the nitrite ion, NO 2 , will influence the pH. Nitrous acid and the
nitrite ion are a conjugate acid-base pair of a weak acid, and the presence of both
in the solution makes the solution a buffer. We need to finish the stoichiometry
part of the problem by determining the molarity of each of these substances.
M HNO2 a 0.060 M HNO2 0.0075 mol HNO2 mL
ba1000 b
(100.00 25.00) mL L
0.00750 mol NO 2
mL
M NO2 a
ba1000 b 0.0600 M NO2
(100.00 25.00) mL L
We can now use these two values for the equilibrium portion of the problem.
There are two options for this buffer solution. We can use these concentrations
in a K a calculation, or we can use the Henderson-Hasselbalch equation. Either
method will give you the same answer; however, the Henderson-Hasselbalch
equation is faster.
The Henderson-Hasselbalch equation is:
pH pK a log
[A ]
[HA]
The pK a appears in the problem (3.35), HA is nitrous acid, and A is the nitrite
ion. We can now enter the appropriate values into this equation:
[0.0600]
pH 3.35 log 3.35
[0.060]
When the titration is at the halfway point (concentration of the conjugate acid
equals the concentration of the conjugate base), the pH will equal the pK a, and
the pOH will equal the pK b.
This is the pH for this point in the second region. All other calculations in this
region work this way.