5 Steps to a 5 AP Chemistry

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

is the limiting reagent.

The stoichiometry part of the problem is finished.

The solution is no longer HNO 2 and NaOH, but HNO 2 and NO 2 −(a conjugate acid–

base pair).

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 (0.00750/0.00750) =3.35

Note the simplification in the CB/CA concentrations. Both moles are divided by exactly

the same volume (since they are in the same solution), so the identical volumes cancel.

c. 49.50 mL. Since both an acid and a base are present (and they are not conjugates), this

must be a stoichiometry problem again. Stoichiometry requires a balanced chemical

equation and moles.

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

is the limiting reagent.

The stoichiometry part of the problem is finished.

The solution is no longer HNO 2 and NaOH, but HNO 2 and NO 2 −(a conjugate acid–

base pair).

HNO NaOH Na NO H O

init. 0

22 +→++ 2

+−

..0150 0.0148 mol 0 0

react.. −−0.0148 0.0148 +0.0148 +0.0 1148

final 0.0002 0.000 — 0.0148

Base:

0.300 mol

1000 mL

49.50 mL = 0.0148 mole

HNO 22 +→++NaOH Na+ NO– H O 2

0.00750 mol base

0 12500Lsolution

0 00750

.

.

⎡

⎣

⎢

⎤

⎦

⎥

mmol acid

0.12500 L solution

⎡

⎣⎢

⎤

⎦⎥

HNO 2 +→ +NaOH Na+ NO H O

init. mol

react.

22

0 0150 0 00750 0 0

0

−+

−

..

.. 0148 0 00750. 0 00750. 0 00750.

fi

−++

nnal 0.00750 0.000 — 0.00750

Base:

0.300 mol

mL

mL = 0.00750 mol

1000

25 00.

Acid:0.150 mol

mL

mL = 0.0150 mol (T

1000

100 00.hhis number will be used in all remaining steps..)

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