Module 8

Equilibrium in Acid-Base Systems


Introduction


In Module 7, you learned about chemical equilibrium. In this module, you will utilize your understanding of the principles of equilibrium to better understand chemical systems containing aqueous acids and bases.



Module 8 contains four lessons:

Lesson 1
Ionization of Water and pH/pOH
Lesson 2
Bronsted-Lowry Theory
Lesson 3
Acid/Bases and the Equilibrium Law
Lesson 4
pH Curves and Buffers



This module builds on your knowledge of acids and bases gained from previous chemistry courses. You may wish to review the following concepts in the textbook.

  • The nature of acidic and basic solutions, page 236

  • The hydronium ion, pages 237 and 248 - 249

  • pH scale (figure 3), page 239

  • pH and pOH calculations, pages 240 - 244

  • Acid-base indicators, page 245

  • The strength of acids and bases, pages 254 - 257

  • Polyprotic substances, page 258

  • Titration of acids and bases, pages 328-330

  • Titration curves, pages 333-336

Module 8 requires competency in pH and pOH calculations. To review this subject, please watch the following instructional video and complete the self-check questions below.


Watch

 
 


Check Your Understanding


Practice Questions 7 and 8 on page 718.

Page 718 Practice Question 7

Food [H3O+(aq)]
(mol/L) 		[OH-(aq)]
(mol/L) 		pH
(mol/L) 		pOH
(mol/L)
Oranges \( \mathrm { 5.5 \times 10^{-3 } } \) \( \mathrm { 1.8 \times 10^{-12} } \)
\( \mathrm { \left( = \dfrac{1.0 \times 10^{-14} }{ 5.5 \times 10^{-3} } \right) } \) 		\( \mathrm { 2.26 } \)
\( \mathrm { ( = - log (5.5 \times 10^{-3}) } \) 		\( \mathrm { 11.74 } \)
\( \mathrm { ( = 14.0 - 2.26) } \)
Asparagus \( \mathrm { 4 \times 10^{-9} } \)
\( \mathrm { \left( = \dfrac{1.0 \times 10^{-14}}{ 3 \times 10^{-6}} \right) } \) 		\( \mathrm { 3 \times 10^{-6} } \)
\( \mathrm { (=10^{-5.6}) } \) 		\( \mathrm { 8.4 } \)
\( \mathrm { (= 14.00 - 5.6) } \) 		\( \mathrm { 5.6 } \)
Olives \( \mathrm { 5 \times 10^{-4} } \)
\( \mathrm { \left( = \dfrac{ 1.0 \times 10^{-14} }{ 2.0 \times 10^{-11}} \right) } \) 		\( \mathrm { 2.0 \times 10^{-11 } } \) \( \mathrm { 3.30 } \) \( \mathrm { 10.70 } \)
\( \mathrm { ( = 14.00 - 3.30) } \)
Blackberries \( \mathrm { 4 \times 10^{-4} } \)
\( \mathrm { (=10-3.4) } \) 		\( \mathrm { 3 \times 10^{-11} } \)
\( \mathrm { (=10-10.6) } \) 		\( \mathrm { 3.4 } \)
\( \mathrm { ( = 14.00-10.6) } \) 		\( \mathrm { 10.6 } \)

Page 718 Practice Question 8

NaOH(aq) → Na+(aq) + OH-(aq)

Because NaOH(aq) is highly soluble in water, it dissociates completely.

Therefore, the [OH-(aq)] = [NaOH(aq)].

\( \mathrm { \text{ Moles of NaOH(aq) } = \dfrac{ 26 g }{ \left(22.99 \frac{g}{mol} + 1.01 \frac{g}{mol} + 16.00 \frac{g}{mol} \right)} = 0.65~ mol } \)

\( \mathrm { \text{ Concentration of NaOH(aq)} = \dfrac{0.65 ~mol}{ 0.150~ L} = 4.3~ \frac{mol}{L } } \)

\( \mathrm { [OH^-(aq)] = 4.3 \frac{mol}{L} } \)

\( \mathrm { pOH = -log (4.3 \frac{mol}{L}) = -0.64 } \)

\( \mathrm { pH = 14.00 - (-0.64) = 14.64 } \)