Chemical Equilibrium and Concentration

Chapter 16: 1-4 Homework

• Reading Preparation
• WebLecture
• Study Activity
• Preparation work for chat
• Online Quiz
• Lab Instructions

Reading Preparation

Textbook assignment: Read Kotz and Triechel, Chemistry and Chemical Reactivity Chapter 16: Sections 1 to 4.

Study Notes

There are three common definitions of acids and bases, the Arrhenius theory, the Brønsted-Lowry theory, and the Lewis theory. In all three, acids donate or promote the concentrations of hydrogen ions (H⁺ protons). Bases may accept these ions, or promote the concentration of hydroxide ions (OH^-) .

1. 16.1 The Arrhenius definition describes an acid as any substance which, when dissolved in water, increases the H+ ion concentration. The Brønsted-Lowry concept identifies acids as proton donors. In Brønsted-Lowry terms, hydrochloric acid HCl acts as a proton donor. Some molecules, such as H₂SO₄ have multiple protons which they can donate, and so are polyprotic acids.
   When an acid and base interact, the resulting products are also an acid-base pair. If HCO₃- reacts to accept a proton, it is a base, but the H₂CO₃ molecule it forms is an acid capable of donating an H+ in the reverse reaction.
2. 16.2 The reaction for dissociation of water molecules is $$2H_{2}O\to {H_{3}O}^{+}+{\mathrm{OH}}^{-}$$
   The equilibrium constant K_w thus equals $$\frac{[{H_{3}O}^{+}][{\mathrm{OH}}^{-}]}{{{[H}_{2}O]}^2}$$ but our rules define [H₂O] = 1.

   So K_w simplifies to $$K_w=[{H_{3}O}^{+}][{\mathrm{OH}}^{-}]$$ which experimentally is 1 * 10^-14.

   In a neutral solution, [H₃O⁺] = [OH^-], so the concentrations at neutralization must be 10^-7 to be equal and give K_w = 1 * 10^-14. If we express the concentrations [H₃O⁺] as 10^-a and [OH^-] as 10^-b, a + b = 14. Increasing b decreases a. Chemists use these exponents as shorthand for the concentration of acids and bases: $$\begin{array}{c}\mathrm{pH}=a=-\log [{H_{3}O}^{+}]\\ \mathrm{pOH}=b=-\log [{\mathrm{OH}}^{-}]\end{array}$$

3. 16.4 In a acid reaction where HA+ H₂O → A^- + H₃O⁺, the equilibrium constant will be $$K_a=\frac{[{H_{3}O}^{+}][A^{-}]}{[\mathrm{HA}]}$$
   In a base reaction where B + H₂O → BH⁺ + OH^-, the equilibrium constant will be $$K_b=\frac{[{\mathrm{BH}}^{+}][{\mathrm{OH}}^{-}]}{[B]}$$
   How much an acid dissociates determines whether it is a weak acid (there is only a small change in HA concentration) or a strong acid (the HA concentration drops to near zero). The strength of the conjugate base will be inversely related to the strength of the acid: the stronger the acid, the weaker the conjugate base.
4. 16.4 In an acid-base reaction, the cation of the base and the anion of the acid form a salt, and the donated H+ and OH^- form neutral water: NaOH + HCl → NaCl + H₂O.

Web Lecture

Read the following weblecture before chat: Acids and Bases Revisited

Study Activity

Chat Preparation Activities

• Essay question: The Moodle forum for the session will assign a specific study question for you to prepare for chat. You need to read this question and post your answer before chat starts for this session.
• Mastery Exercise: The Moodle Mastery exercise for the chapter will contain sections related to our chat topic. Try to complete these before the chat starts, so that you can ask questions.

Chapter Quiz

• There is no chapter quiz YET.

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(Aligns to) AP #12 GUIDED INQUIRY — Determining the rate law — Phase III

Complete your data analysis and submit a formal report of your results.

Resources:

• IGHCE Lab 12.3 Determination of the effect of Concentration on Reaction Rate
• HSCKM VI-3 Determining a Reaction Order
• APGIE Investigation 11: What Is the Rate Law of the Fading of Crystal Violet Using Beer’s Law
• There is no alternative form of this lab.

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