Untitled - Kelly Walsh High School

248 CHEMISTRY FOR THE UTTERLY CONFUSED
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(55.00 mL)
L 1000 mL
0.0165 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 0.0165 0 0 —
Reacted moles 0.0150 0.0150 0.0150 0.0150
Postreaction 0.0000 0.0015 0.00150 0.0150
There are two bases present after the reaction. Both of these could influence
the pH. However, since the sodium hydroxide is a strong base, it will be more
important than the weaker base, the nitrite ion. We need to determine the sodium
hydroxide ion concentration after the reaction.
0.0015 mol NaOH mL
M NaOH a ba1000
(100.00 55.00)mL 1L b
9.677 10 3 M NaOH (unrounded)
Sodium hydroxide is a strong base so M NaOH M OH . We can use the
hydroxide ion concentration to determine the pH:
pOH log [OH ] log (9.677 10 3 ) 2.01426 (unrounded)
pK w pH pOH 14.00
pH pK w pOH 14.00 2.01426 11.98574 11.99
All other calculations in this region work in a similar manner.