practice solutions - 05 - acid-base equilibrium - strong and weak bases

7/23/2019 Practice Solutions - 05 - Acid-Base Equilibrium - Strong and Weak Bases

1/4

Advanced Chemist ry Pract ice Prob lems

Strong and Weak Bases

1. Question: Identify each of the following as a strong or weak base.

a. KOH

b. NH3

c. CH3NH2

d. Ba(OH)2

Answer: Strong bases contain the hydroxide anion while most weak bases will be

an ammine compound (contain an N as a central atom).

a. KOHstrong

b. NH3weak

c. CH3NH2

weakd. Ba(OH)2strong

2. Question: Complete the table. Assume all solutions are at 25C.

Solution [H+] pH [OH-] pOH

HBr 0.751

HF 3.2 10-4

HClO4 1.04

CH2O2 5.35 10-10

NaOH 13.54

NH3 4.35KOH 2.53 10-2

Answer: If one of the four values ([H+], [OH-], pH or pOH) is known, then the

other three values can be calculated. The value of Kwat the given temperature

must be known. In these examples, the temperature is 25C and the value of

Kwis 1.010-14.

Solution [H+] (M) pH [OH-] (M) pOH

HBr 0.177 0.751 5.65 10-14 13.249

HF 3.2 10-4 3.49 3.1 10-11 10.51HClO4 0.091 1.04 1.10 10-13 12.96

CH2O2 1.87 10-5 4.728 5.35 10-10 9.272

NaOH 2.9 10-14 13.54 0.46 0.34

NH3 2.2 10-10 9.65 4.5 10-5 4.35

KOH 3.95 10-13 12.403 2.53 10-2 1.597
http://upload.wikimedia.org/wikipedia/commons/e/e5/Coursera_logo.PNG


2/4


To convert pH to [H+] [H+] = 10-pH

To convert [H+] to pH pH = -log[H+]

To convert pH to pOH pOH = 14 - pH

To convert pOH to pH pH = 14 - pOH

To convert pOH to [OH-] [OH] = 10-pOH

To convert [OH-] to pOH pOH = -log[OH-]

To convert between [H+] and [OH-] Kw= [H+][OH-]

at 25C 1.0 10-14 = [H+][OH-]

1. Question: Determine the pH for each of the given solutions.

a. 0.250 M KOH

b. 0.250 M Ba(OH)2

c. 0.250 M NH3, Kb= 1.8 10-5

Answer: To find the pH of these basic solutions, the pOH needs to be calculated

first. The approach will be different for strong and weak bases.

a. KOH is a strong base. For 0.250 M KOH, the [OH-

] = [KOH] therefore [OH-

] = 0.250 M

0.602pOH

]log[0.250pOH

]log[OHpOH -

With the value of the pOH, the pH can be determined by subtracting

from 14.

13.398pH

0.60214pH

pOH14pH

14pOHpH

Given that KOH is a strong base, a high pH value (close to 14) is

expected.


3/4


b. For 0.250 M Ba(OH)2, the approach will be similar except for the fact that

there are two hydroxide groups for every unit of Ba(OH)2so the hydroxideconcentration is twice that of the base. Therefore [OH-] = 0.500 M

0.301pOH

]log[0.500pOH

]log[OHpOH -

With the value of the pOH, the pH can be determined by subtracting

from 14.

13.699pH

0.30114pH

pOH14pH

14pOHpH

Given that the hydroxide concentration is higher for Ba(OH)2than for

KOH, the pH will be higher (more basic).

c. 0.250 M NH3, Kb= 1.8 10-5

Ammonia is a weak base. Therefore, an equilibrium equation and ICE

table must be used to determine the pOH, and then the pH of the solution.

The Kbvalue (givn in the problem) is 1.8 10-5.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

INITIAL: The initial concentration of NH3is given as 0.250 M. The initial

concentrations of NH4+and OH-are zero and the concentration of water is

omitted.

CHANGE: The change is represented by x and follows the stoichiometry

of the balanced chemical reaction.

EQUILIBRIUM: The equilibrium row is the sum of the initial and change

rows for each column.


4/4


NH3 (M) H2O (M) NH4+(M) OH-(M)

I 0.250 M 0 0C -x +x +x

E 0.250 - x x x

The law of mass action is given and the equilibrium values may be

substituted into the equation to solve for x and thus the value of [OH -].

x0.250xx

101.8

NH

OHNH

5-

3

-

4

b

-

K

The value of Kbis small, so the assumption that x

practice solutions - 05 - acid-base equilibrium - strong and weak bases

Documents