Electrolytes and Acid-Base Reactions

Note on Solubility: Do not confuse solubility with electrolyte strength. For example, $\text{BaSO}_4$ has very low solubility, but the tiny amount that does dissolve dissociates completely into $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ ions, making it a strong electrolyte.

Weak Electrolytes

Weak electrolytes only partially dissociate into ions. Instead of breaking apart completely, they exist in a state of chemical equilibrium where the unionized molecules and the produced ions coexist. Because the concentration of ions is low, these solutions are only slightly conductive.

A classic example is acetic acid (vinecous acid), which can be represented by the following equilibrium equation: $\text{CH}_3\text{CO}_2\text{H}(aq) \rightleftharpoons \text{CH}_3\text{CO}_2^-(aq) + \text{H}^+(aq)$

Non-Electrolytes

Non-electrolytes are substances that undergo no ionization upon dissolution. While they may dissolve completely in water, they remain as neutral molecules rather than forming ions, and therefore cannot conduct electricity. Examples include sugar ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$), ethanol, and ethylene glycol.

Brønsted–Lowry Acid-Base Theory

Once we understand how ions behave, we can look at specific types of reactions involving hydrogen ions ($\text{H}^+$). The Brønsted–Lowry theory defines acids and bases based on the transfer of protons.

• Brønsted–Lowry Acids: These are proton donors—species that give away an $\text{H}^+$ ion.
• Brønsted–Lowry Bases: These are proton acceptors—species that receive an $\text{H}^+$ ion.

The Hydronium Ion

In aqueous solutions, a "free" hydrogen ion ($\text{H}^+$) does not actually exist in isolation. Because $\text{H}^+$ is highly reactive, it immediately associates with a water molecule ($\text{H}_2\text{O}$) to form the hydronium ion ($\text{H}_3\text{O}^+$).

Conjugate Acid-Base Pairs

When an acid donates a proton, it is transformed into a different species. This relationship creates conjugate acid-base pairs.

1. The conjugate base is what remains after the acid has lost its proton.
2. The conjugate acid is what is formed after the base has accepted a proton.

Consider the reaction between hydrochloric acid and water: $\text{HCl}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{O}^+(aq) + \text{Cl}^-(aq)$ In this reaction, $\text{HCl}$ is the acid (donor) and $\text{H}_2\text{O}$ is the base (acceptor). Consequently, $\text{Cl}^-$ is the conjugate base of $\text{HCl}$, and $\text{H}_3\text{O}^+$ is the conjugate acid of $\text{H}_2\text{O}$.

Exam Tip: Water is an amphiprotic substance, meaning it can act as either a proton donor or a proton acceptor depending on what it is reacting with.

Neutralization and Titration

When a Brønsted–Lowry acid reacts with a Brønsted–Lowry base, a neutralization reaction occurs, typically producing a salt and water. For example: $\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$

Titration: Measuring Concentration

A titration is a volumetric method used to determine the concentration of an unknown substance by reacting it with a standard solution (a solution of known concentration).

During a titration, we look for two key moments:

• Equivalence Point: The point where enough titrant has been added to completely react with the unknown substance.
• End Point: The point where a chemical indicator changes color, signaling that the reaction is complete.

Worked Example: Stoichiometry from Titration

Problem: What is the concentration of sulfuric acid ($\text{H}_2\text{SO}_4$) if $15.00\text{ mL}$ of it reacts with $18.45\text{ mL}$ of a $0.0973\text{ M NaOH}$ solution?

Step 1: Write the balanced equation $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Step 2: Calculate moles of base ($\text{NaOH}$) $\text{moles}_{\text{base}} = \text{Molarity} \times \text{Volume (L)}$ $\text{moles}_{\text{base}} = 0.0973\text{ M} \times 0.01845\text{ L} = 0.00180\text{ mol}$

Step 3: Use stoichiometry to find moles of acid From the equation, $1\text{ mole of acid}$ reacts with $2\text{ moles of base}$. $\text{moles}_{\text{acid}} = 0.00180\text{ mol base} \times \frac{1\text{ mol acid}}{2\text{ mol base}} = 0.0009\text{ mol acid}$

Step 4: Calculate molarity of the acid $\text{M}_{\text{acid}} = \frac{0.0009\text{ mol}}{0.015\text{ L}} = 0.060\text{ M}$