5 Steps to a 5 AP Chemistry

Since a CA/CB pair is present, this is now a buffer problem, and the Henderson–
Hasselbalch equation may be used.

pH =pKa+log (CB/CA) =3.35 +log (0.0148/0.0002) =5.2

d. 50.00 mL. Since both an acid and a base are present (and they are not conjugates), this
must be a stoichiometry problem. Stoichiometry requires a balanced chemical equation
and moles.

Based on the stoichiometry of the problem, and on the moles of acid and base, both
are limiting reagents.

The stoichiometry part of the problem is finished.
The solution is no longer HNO 2 and NaOH, but an NO 2 – solution (a conjugate base
of a weak acid).
Since the CB of a weak acid is present, this is a Kbproblem.

e. 55.00 mL. Since both an acid and a base are present (and they are not conjugates), this must
be a stoichiometry problem. Stoichiometry requires a balanced chemical equation and moles.

Based on the stoichiometry of the problem, and on the moles of acid and base, the acid
is now the limiting reagent.

HNO NaOH Na NO H O

init. 0

22 +→+ + 2

+−

..0150 0.0165 mol 0 0

react.. −−0.0150 0.0150 +0.0150 +0.0 1150

final 0.0000 0.000 — 0.0150

HNO NaOH Na NO H O

Base:

0.300 mol

mL

222

+→+++ −

1000

55.000 mL = 0.0165 mol

K

xx

x

x

x

b==×=− neglect

=

10 −−2 24 10

0 100

10 65.. 11 ()()

.

[OOH M p OH

pH pOH

−−

−

=× =

=− =−

]..

...

150 10 582

14 00 14 00 5 8

6

22818 =.

NO 2 H O 22 OH H NO

0 100

• 
  

.

++

−

 −

xxx

ppKKba=−=−=14 000....14 000 3 35 10 65

HNO NaOH Na NO H O

init. 0

22 +→+ + 2

+−

..0150 0.0150 mol 0 0

react.. −−0.0150 0.0150 +0.0150 +0.0 1150

final 0.0000 0.000 — 0.0150

[].NO 2 =0 0150mol/.0 150LLM=0 100.

Base:

0.300 mol

mL

mL = 0.0150 mol

1000

50 00.

HNO 22 +→++NaOH Na+ NO 2 – H O

Equilibrium  227