5 Steps to a 5 AP Chemistry

Let’s consider a typical titration problem. A 100.0 mL sample of 0.150 M nitrous acid
(pKa= 3.35) was titrated with 0.300 M sodium hydroxide. Determine the pH of the
solution after the following quantities of base have been added to the acid solution:

a. 0.00 mL
b. 25.00 mL
c. 49.50 mL
d. 50.00 mL
e. 55.00 mL
f. 75.00 mL

a. 0.00 mL. Since no base has been added, only HNO 2 is present. HNO 2 is a weak acid,
so this can only be a Kaproblem.

Quadratic needed: x^2 +4.47 × 10 −^5 x−6.70 × 10 −^5 = 0

(extra sig. figs.)

b. 25.00 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.

Na++NO 2 −could be written as NaNO 2 , but the separated ions are more useful.

HNO 2 +→++NaOH Na+ NO− 2 H O 2

x= [H ] = 8.0 10 M pH = 2.10+ × −^3

K

xx

a x

()

==×=

−

10 −−4 5 10

0 150

335.. 4 ()

.

HNO 22 H aq) + NO+(

.

−

0 150−xx x

Equilibrium  225

14

12

10

8

6

4

2

0

pH

10 20 30 40 50 60 70 80

Volume of NaOH added (mL)

Figure 15.3 The Titration of a weak acid with a strong base.