7.1BRONSTED-Lowry Acids, Bases, and the Ion Product of Water

Section Learning Objectives

• Identify substances as acids or bases according to the Brønsted-Lowry definition.

• Identify conjugate acid-base pairs within chemical reactions.

• Write balanced chemical equations representing acid and base ionization reactions.

• Utilize the ion product constant ($K_w$) to calculate the concentrations of hydronium ($H_3O^+$) and hydroxide ($OH^-$) ions.

• Describe the chemical behavior of amphiprotic substances, which can function as both acids and bases.

The pH Scale and Buffer Solutions

• The pH scale is a color-coded measure of the hydrogen ion concentration in a system.

• Acidic Solutions: Systems with a pH between $0$ and $7$ are characterized by a higher concentration of hydrogen ions.

• Neutral Solutions: When the concentration of hydrogen ions is equal to the concentration of hydroxide ions, the solution is neutral and has a pH of exactly $7$.

• Basic (Alkaline) Solutions: Systems with very low quantities of hydrogen ions and higher quantities of hydroxide ions have a pH between $7.1$ and $14$.

• Buffer: A buffer is a solution capable of resisting changes in pH when small quantities of an acid or a base are added to it.

Brønsted-Lowry Acid-Base Definitions

• The Brønsted-Lowry definitions are more broadly applicable than previous definitions, such as the Arrhenius definitions.

• Brønsted-Lowry Acid: A substance that donates a proton ($H^+$).

• Brønsted-Lowry Base: A substance that accepts a proton ($H^+$).

• In these reactions, a proton is transferred from the donor (acid) to the acceptor (base).

Conjugate Acid-Base Pairs

• A conjugate pair consists of the two species between which a proton is transferred.

• Conjugate Base: This is the species that remains after an acid has donated its proton.

• Conjugate Acid: This is the species that remains after a base has accepted a proton.

• Example: Interaction of Water and Ammonia

  • Reaction:   $H_2O + NH_3 \rightleftharpoons NH_4^+ + OH^-$

  • In this scenario, water ($H_2O$) acts as the Brønsted-Lowry acid because it donates a proton to ammonia ($NH_3$).

  • Ammonia ($NH_3$) acts as the Brønsted-Lowry base.

  • Ammonium ($NH_4^+$) is the conjugate acid formed from the base ($NH_3$).

  • Hydroxide ($OH^-$) is the conjugate base formed from the acid ($H_2O$).

  • In the reverse reaction, the conjugate acid ($NH_4^+$) would donate a proton back to the conjugate base ($OH^-$) to reform the initial reactants.

Acid and Base Ionization

• Acid Ionization: This refers to the event where neutral particles dissociate or dissolve in water to produce charged particles.

  • Example: Hydrogen fluoride ($HF$) dissolved in water.

  • Reaction:   $HF + H_2O \rightleftharpoons H_3O^+ + F^-$

  • $HF$ behaves as the acid by donating a proton to water, which acts as the base.

  • In the reverse reaction, hydronium ($H_3O^+$) would donate a proton back to fluoride ($F^-$).

• Base Ionization: This occurs when a substance accepts a proton from water. In base ionization reactions, water functions as the acid.

  • Example: Pyridine molecule reacting with water.

  • The pyridine molecule accepts a proton from water, which moves toward the nitrogen in the pyridine.

  • This forms a conjugate acid (the protonated pyridine) and a conjugate base ($OH^-$).

  • In the reverse reaction, the proton moves from the conjugate acid back to the hydroxide.

Amphiprotic and Amphoteric Substances

• Definition: Substances that can either donate or accept a proton are described as amphiprotic or amphoteric.

• Example: Bicarbonate (Hydrogen Carbonate, $HCO_3^-$)

  • Bicarbonate consists of a hydrogen ion and a carbonate ion. When dissolved in water, it can participate in two simultaneous equilibria.

  • Reaction 1 (Acting as an Acid):   $HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$

  • Here, bicarbonate donates a proton to water, forming the conjugate base carbonate ($CO_3^{2-}$) and the conjugate acid hydronium ($H_3O^+$).

  • Reaction 2 (Acting as a Base):   $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$

  • Here, bicarbonate accepts a proton from water, forming the conjugate acid carbonic acid ($H_2CO_3$) and the conjugate base hydroxide ($OH^-$).

• These processes occur simultaneously and equally in the solution to establish equilibrium.

Autoionization of Water

• Water itself is amphoteric and can undergo a process called autoionization.

• In autoionization, like molecules react to produce ions: one water molecule acts as an acid (donating a proton) while another acts as a base (accepting the proton).

• Reaction:   $H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$

• Statistical Frequency: Under standard conditions ($25\,^{\circ}C$), autoionization is rare. It is estimated that only $2$ out of every $1,000,000,000$ water molecules undergo this process.

• Temperature Effects: Increasing the temperature increases the amount of autoionization. For example, moving from $25\,^{\circ}C$ to $100\,^{\circ}C$ will result in more autoionization.

The Ion Product Constant for Water ($K_w$)

• The equilibrium constant for the autoionization of water is known as the ion product constant, denoted as $K_w$.

• Because water is the solvent, it is not included in the denominator of the equilibrium expression.

• The expression is defined as the product of the concentrations of the hydronium ion and the hydroxide ion:   $K_w = [H_3O^+][OH^-]$

• At $25\,^{\circ}C$, the value of $K_w$ is exactly:   $K_w = 1 \times 10^{-14}$

• Phase labels: The ions ($H_3O^+$ and $OH^-$) are labeled as aqueous ($aq$), while the reacting water is liquid ($l$).

• This constant allows for the calculation of an unknown ion concentration if the other is known.

Guided Practice: Calculating Ion Concentrations

• Problem 1: Calculating concentration at $80\,^{\circ}C$

  • Given: At $80\,^{\circ}C$, $K_w = 2.4 \times 10^{-13}$.

  • Goal: Solve for the concentration of hydrogen and hydroxide ions in pure water.

  • Since the ions are produced in a $1:1$ ratio, let $x = [H^+] = [OH^-]$.

  • Equation:   $2.4 \times 10^{-13} = x \times x = x^2$

  • Solving for $x$:   $x = \sqrt{2.4 \times 10^{-13}}$

  • Calculation result: $x = 4.898979486 \times 10^{-7}$

  • Applying proper significant figures (underlining the first two digits):   $x = 4.9 \times 10^{-7}\,M$

• Problem 2: Calculating hydronium concentration at $25\,^{\circ}C$

  • Given: Hydroxide concentration $[OH^-] = 0.001\,M$ at $25\,^{\circ}C$.

  • Known: At $25\,^{\circ}C$, $K_w = 1.0 \times 10^{-14}$.

  • Goal: Solve for $[H_3O^+]$.

  • Equation:   $1.0 \times 10^{-14} = [H_3O^+][0.001]$

  • Isolation of unknown:   $[H_3O^+] = \frac{1.0 \times 10^{-14}}{0.001}$

  • Calculation result:   $[H_3O^+] = 1.0 \times 10^{-11}\,M$

Practice: Amphoteric Reactions of Dihydrogen Phosphate

• Substance: Dihydrogen phosphate ($H_2PO_4^-$).

• Scenario 1: Acting as a Base with Hydrogen Bromide ($HBr$)

  • $H_2PO_4^-$ accepts a proton from the acid $HBr$.

  • Adding $H^+$ to $H_2PO_4^-$ neutralizes the $-1$ charge to form phosphoric acid ($H_3PO_4$).

  • The $HBr$ loses a proton to become the bromide ion ($Br^-$).

  • Equation:   $H_2PO_4^-(aq) + HBr(aq) \rightleftharpoons H_3PO_4(aq) + Br^-(aq)$

• Scenario 2: Acting as an Acid with Hydroxide ($OH^-$)

  • $H_2PO_4^-$ donates a proton to the base $OH^-$.

  • Water is formed from the hydroxide accepting the proton ($H^+ + OH^- \rightarrow H_2O$).

  • Dihydrogen phosphate loses a proton, changing its charge from $-1$ to $-2$ to form hydrogen phosphate ($HPO_4^{2-}$).

  • Equation:   $H_2PO_4^-(aq) + OH^-(aq) \rightleftharpoons HPO_4^{2-}(aq) + H_2O(l)$

  • Note: $H_2O$ is in the liquid phase as it is the solvent.