5 Steps to a 5 AP Chemistry 2019

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

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 Ka problem.

 +

−

+ −

xx x

HNO H(aq)NO

0.150

22

==×=

−

K −−xx

x

10 4.5 10

()()

a 0.150

3.35 4

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

(extra sig. figs.)

x==[H ] 8.0+ ×=10 Mp−^3 H 2.10

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.

HNO 22 +→NaOH Na+−++NO HO 2

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

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.