5 Steps to a 5 AP Chemistry 2019

228 ❯ STEP 4. Review the Knowledge You Need to Score High

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.

HNO 22 +→NaOH Na+++NO– HO 2

Base: =

0.300mol

1,000 mL

50.00mL 0.0150mol

Based on the stoichiometry of the problem, and on the moles of acid and base, both

are limiting reagents.

+→++

−− ++

==

+−

M

HNONaOHNaNOHO

init. 0.0150 0.0150mol0 0

react. 0.0150 0.0150 0.0150 0.0150

final 0.0000 0.000 — 0.0150

[NO]0.0150mol/0.150L 0.100

222

2

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 Kb problem.

pKKba=−14.000 p =−14.000 3.35=10.65

++

−

−

xxx

NOHOOHHNO

0.100

2

• 22

K

xx

x

x

x

10 2.24 10

()()

0.100

(neglect )

[OH] 1.50 10 MpOH 5.82

pH 14.00 pOH 14.00 5.82 8.18

b

10.65 11

6

==×=

−

−

==×=

=−=−=

−−

−−

−

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.

+→++

=

HNO NaOH Na+ NO− HO

Base:

0.300mol

1,000 mL

55.00mL 0.0165mol

222

Based on the stoichiometry of the problem, and on the moles of acid and base, the

acid is now the limiting reagent.

HNONaOHNaNOHO

init. 0.0150 0.0165 mol0 0

react. 0.0150 0.0150 0.0150 0.0150

final 0.0000 0.0015 — 0.0150

22 +→++ 2

−− ++

+−