Module 8 

Lesson 1.2  Acid Strength



Key Concepts


Acids can be classified as strong or weak depending on their ability to produce hydronium ions, a process referred to as ionization. A strong acid has a greater tendency to lose a proton than does a weak acid.


Acid strength has nothing to do with concentration. Concentration refers simply to the chemical quantity of acid per unit volume of solution.


For example, HCl is considered a strong acid. If you dilute an HCl solution, it becomes less concentrated. However, dilution does not change the fact that HCl reacts quantitatively with water to form hydronium ions.


There are six strong acids: hydrochloric, nitric, sulfuric, hydrobromic, hydroiodic, and perchloric.


Because all six strong acids react quantitatively with water to form hydronium ions, the hydronium ion concentration of any strong acid at equilibrium is equal to the initial concentration of the acid dissolved in the aqueous solution.


\( \mathrm { HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq) } \)


Although a quantitative equation can be written with double equilibrium arrows and a 99.9% note, a single arrow (→) is more commonly used to indicate that the reaction is virtually complete.


Fig. 1


The hydronium ion is the strongest acid that can exist in aqueous solution. If any acid stronger than the hydronium ion is dissolved in water, it reacts instantly with water molecules to form hydronium ions. Therefore, all six strong acids are considered equivalent in strength.


Weak acids vary in strength because they ionize in water to various extents. To construct the table entitled "Relative Strengths of Aqueous Acids and Bases", solutions of acids at the same concentration were compared, based on their extent of ionization (as measured by pH). Weak acids only partially ionize to form hydronium ions. In fact, most weak acids ionize less than 50%.


Acetic acid is an example of a weak acid.


Fig. 2

Recall that any equilibrium position depends on concentration as well as on temperature. 1.3% is valid only for a 0.10 mol/L solution at 25°C.


Because only a small proportion of the initial acid concentration will be converted to ions at equilibrium, the hydronium ion concentration at equilibrium is not equal to the initial amount concentration of the weak acid solution. For weak acids, the hydronium ion concentration is equal to the initial concentration of the acid solution multiplied by its percent ionization (1.3%). Refer to Communication example 5 on page 720 to see how to perform this calculation.


Once the hydronium concentration of a weak acid is known, the pH of the solution can be calculated using pH = -log[H3O+]. Remember that the number of digits following the decimal in the pH or pOH value should equal the number of significant digits in the amount concentration of the ion. Refer to the Learning Tip on page 718 of the textbook.


It is possible to calculate percent ionization of a weak acid when you are provided with the pH of that acid. Work through Communication examples 6 to see how the percent ionization of a weak acid can be determined using pH.


Fig. 3


Watch


 



Important!

Remember that pH is a measure of H3O+(aq) concentration and not acid strength. It is possible for a highly concentrated weak acid to have a lower (more acidic) pH than a dilute strong acid. This is the reason why acid solutions are always compared at the same concentration!

 Read pages 718-720 in the textbook.


Check Your Understanding


Complete Section 16.1 questions 4, 6, and 7 on page 721 of your textbook.

Page 721 Section 16.1 Question 4



  1. \( \mathrm { [H3O^+(aq)] = 0.10\frac{mol}{L} \left( \dfrac{7.8 \times 10^{-3}}{100} \right) = 7.8 \times 10^{-6} \frac{mol}{L} } \)


    \( \mathrm { pH = -log (7.8 \times 10^{-6} \frac{mol}{L}) = 5.11 } \)


Page 721 Section 16.1 Question 6

\( \mathrm { [H3O^+(aq)] = \dfrac{1.0 \times 10^{-14} }{ 2.5 \times 10^{-7} \frac{mol}{L} } = 4.0 \times 10^{-8} \frac{mol}{L} } \)


\( \mathrm { pH = -log (4.0 \times 10^{-8} \frac{mol}{L}) = 7.40 } \)


Page 721 Section 16.1 Question 7

\( \mathrm { [H3O^+(aq)] = 10^{-5.6} = 3 \times 10^{-6} \frac{mol}{L} } \)