5 Steps to a 5 AP Chemistry 2019

Equilibrium ❮ 227

Acid: =

0.150mol

1,000 mL

100.00mL 0.0150mol

(This number will be used in all remaining steps.)

Base: =

0.300mol

1,000 mL

25.00mL 0.00750mol

(^) Based on the stoichiometry of the problem, and on the moles of acid and base,
NaOH is the limiting reagent.
HNONaOHNaNOHO
init.mo
2 22
0 0150 0 00750

+→+−++

..ll

react.

fi

00

−−0 0148..0 00750 ++0 00750..0 00750

nnal 0.00750 0.000 —0. 00750

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.

0.00750molbase

0.12500Lsolution

0.00750molacid

0.12500Lsolution





















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.

HNO 22 +→NaOH Na+++NO– HO 2

Base: =

0.300mol

1000 mL

49.50mL 0.0148mole

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

NaOH is the limiting reagent.

+→++

−− ++

HNONaOHNaN+−OHO

init. 0.0150 0.0148mol0 0

react. 0.0148 0.0148 0.0148 0.0148

final 0.0002 0.000 — 0.0148

222

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).