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Acids and Bases: Brønsted-Lowry Theory and Equilibria, Study Guides, Projects, Research of Chemistry

An overview of the brønsted-lowry theory of acids and bases, including definitions, examples, and reactions. It covers the concepts of conjugate acids and bases, acid-base equilibria, and strong and weak acids and bases. Students will learn how to identify acids and bases, determine conjugate species, and calculate equilibrium constants.

Typology: Study Guides, Projects, Research

2021/2022

Uploaded on 09/27/2022

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Worksheet 18 - Acids and Bases
The Brønsted-Lowry definition of an acid is a substance capable of donating a proton
(H+), and a base is a substance capable of accepting a proton. For example, the
weak acid, HF, can be dissolved in water, giving the reaction:
HF (aq) + H2O (l) ' H3O+ (aq) + F- (aq)
acid conjugate base
base conjugate acid
In this reaction, HF is the species losing the proton (H+), making it the acid. Water is
the species accepting the proton, to form the hydronium ion, H3O+, making it the base.
The F- (aq) is called the conjugate base of HF. It can gain a proton in the reverse
reaction. H3O+ is the conjugate acid of H2O, since it can lose a proton in the reverse
reaction. The stronger an acid, the weaker its conjugate base will be and the stronger
the base, the weaker its conjugate acid.
The equilibrium concentrations of these species will be determined by the relative
strengths of the acids and bases. The strongest acid will dissociate to the greatest
extent. H3O+ (H+) is the strongest acid that can exist in an aqueous system. So, the
equilibrium in this system will favor the reactants, HF and H2O, the weaker acid and
base. The equilibrium state is described by an equilibrium constant, Ka, in the case of
acids, and Kb in the case of bases. These are related by the expression Kw = Ka x Kb =
1 x 10-14.
1. Classify the following as Brønsted acids, bases or both.
If the compound has a hydrogen, it can be an acid; if it has any lone
pairs, it can be a base
a) H2O b) OH- c) NH3
both both both
d) NH4+ e) NH2- f) CO32-
acid both base
2. What is the conjugate base of the following acids?
a) HClO4 b) NH4+ c) H2O d) HCO3-
ClO4- NH
3 OH
- CO
32-
(Remove one hydrogen and reduce the charge by 1)
3. What is the conjugate acid of the following bases?
a) CN- b) SO42- c) H2O d) HCO3-
HCN HSO4- H
3O+ H
2CO3
(Add one hydrogen and increase the charge by 1)
pf3
pf4

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Worksheet 18 - Acids and Bases

The Brønsted-Lowry definition of an acid is a substance capable of donating a proton (H+^ ), and a base is a substance capable of accepting a proton. For example, the weak acid, HF , can be dissolved in water, giving the reaction:

HF (aq) + H 2 O (l) ' H 3 O +^ (aq) + F-^ (aq)

acid conjugate base base conjugate acid

In this reaction, HF is the species losing the proton (H+^ ), making it the acid. Water is the species accepting the proton, to form the hydronium ion , H 3 O +^ , making it the base. The F-^ (aq) is called the conjugate base of HF. It can gain a proton in the reverse reaction. H 3 O +^ is the conjugate acid of H 2 O, since it can lose a proton in the reverse reaction. The stronger an acid, the weaker its conjugate base will be and the stronger the base, the weaker its conjugate acid.

The equilibrium concentrations of these species will be determined by the relative strengths of the acids and bases. The strongest acid will dissociate to the greatest extent. H 3 O +^ (H+) is the strongest acid that can exist in an aqueous system. So, the equilibrium in this system will favor the reactants, HF and H 2 O, the weaker acid and base. The equilibrium state is described by an equilibrium constant, K a , in the case of acids, and K b in the case of bases. These are related by the expression K w = K a x K b = 1 x 10-^.

  1. Classify the following as Brønsted acids, bases or both.

If the compound has a hydrogen, it can be an acid; if it has any lone

pairs, it can be a base

a) H 2 O b) OH-^ c) NH 3

both both both

d) NH 4 +^ e) NH 2 -^ f) CO 3 2-

acid both base

  1. What is the conjugate base of the following acids? a) HClO 4 b) NH 4 +^ c) H 2 O d) HCO 3 -

ClO 4

-

NH 3 OH

-

CO 3

2-

(Remove one hydrogen and reduce the charge by 1)

  1. What is the conjugate acid of the following bases? a) CN-^ b) SO 4 2-^ c) H 2 O d) HCO 3 -

HCN HSO 4 -^ H 3 O+^ H 2 CO 3

(Add one hydrogen and increase the charge by 1)

  1. HSO 3 -^ is amphoteric ; it can behave as either an acid or a base. In the examples above, you may have noticed that H 2 O and HCO 3 -^ can also exist as acids or bases.

a) Write the equation for the reaction of HSO 3 -^ with H 2 O in which it acts like an acid and identify the acid-base pairs. Circle the strongest acid. Then write an expression for K a. The value of K a at 25o^ C is 1.23 x 10-7^. Use an arrow to indicate which side is favored by the equilibrium

HSO 3 -^ + H 2 O ' SO 3 2-^ + H 3 O+

[ ][ ]

[ ]

7 3

3

2 (^3 1). 23 10 − −

− + = = × HSO

SO HO

K (^) a

b) Write the equation for the reaction of HSO 3 -^ with water in which it acts like a base and identify the acid-base pairs. Circle the strongest base. Then write an expression for K b. What is the value of K b at 25o^ C, given that the K a for H 2 SO 3 = 1.58 x 10-2^? Use an arrow to indicate which side is favored by the equilibrium.

HSO 3

-

+ H 2 O ' H 2 SO 3 + OH

-

[ ][ ]

[ −]

3

2 3 HSO

H SO OH

K (^) b

13 2

14

  1. 33 10
  2. 58 10

− = × ×

×

a

w b K

K

K

c) If HSO 3 -^ is placed in water, is the resulting solution acidic or basic?

Ka > Kb , so HSO 3

-

will produce more H 3 O

+

than OH

-

and the

solution overall will be acidic

d) Compute a value for the equilibrium constant for the reaction shown below:

SO 3 2-^ + H 3 O +^ ' HSO 3 -^ + H 2 O K = __ 8.13x10 6 _______

[ ]

[ ][ ]

6 7 3

2 3

fromparta

= ×

×

SO HO K a

HSO

K

Circle the strongest acid and indicate which side is favored by the equilibrium.

  1. Calculate the pH of the solution formed when 100 mL of 1.00 M HCl is added to1.00 L of water.

[ ]

pH -log( 0.0909) 1. 04

0. 0909 M

1. 00 0. 100 L

  1. 100 molH

  2. 100 molH 1 molHCl

1 molH 1 L

  1. 00 molHCl
  2. 100 L

× × =

H

  1. Calculate the pH of the following solutions of strong acids and bases:

a) 0.001 M HCl b) 0.76 M KOH

[ ]

pH -log(0.001 ) 3

0. 001 M

H + =

[ ]

pOH -log0.76 0. 119

0. 76 M

pH

OH

c) 2.8 x 10 -4^ M Ba(OH) 2 (Think carefully about this one!)

When Ba(OH) 2 dissociates, two OH

-

ions are produced, so

the total concentration of OH

-

is double the concentration

of Ba(OH) 2

[ ] ( )

pOH -log 5. 6 10 3. 25

22. 8 10 M 5. 6 10 M

4

4 4

= × =

= × = ×

− − −

pH

OH