Untitled - Kelly Walsh High School

246 CHEMISTRY FOR THE UTTERLY CONFUSED
c. 50.00 mL. This will be in the third region of the titration curve. (We usually
cannot predict this by simple inspection.)
We must again use the reaction:
HNO 2(aq) NaOH(aq) l Na (aq) NO 2 (aq) H2O(l)
We need to deal with the stoichiometry of this reaction. For this reason, we
need to know the moles of each of the reactants. We already found the initial
moles of nitrous acid (0.0150 mol) so we do not need to determine them again.
We can find these from the concentration and the volume of each solution.
0.300 mol NaOH 1L
Moles NaOH a ba b(50.00 mL)
L 1000 mL
0.0150 mol NaOH
We again need to determine the limiting reagent. In this case, both reactants are
limiting.
We will now create a new reaction table:
HNO2(aq) NaOH(aq) l Na (aq) NO2 (aq) H2O(l)
Initial moles 0.0150 0.0150 0 0 —
Reacted moles 0.0150 0.0150 0.0150 0.0150
Postreaction 0.0000 0.00000 0.00150 0.0150
When both reactants are limiting (zero moles remaining), we are at the equivalence
point.
The only substance remaining in the solution that can influence the pH is the
nitrite ion. This ion is the conjugate base of a weak acid. Since a base is present,
the pH will be above 7. The presence of this weak base means this is a Kb problem.
However, before we can attack the equilibrium portion of the problem, we
must finish the stoichiometry part by finding the concentration of the nitrite ion.
0.0150 mol NO
2 mL
M NO2 a
ba1000 0.100 M NO2
(100.00 50.00)mL 1L b
We can now proceed to the equilibrium portion of the problem. To do this we
need the K b. We can find the K b from the following series of relationships
involving the given pK a:
pKw pKa pKb 14.00
pKb pKw pKa 14.00 3.35 10.65