
Chapter 14 Acids and Bases

Shaun Williams, PhD

Models of Acids and Bases

• Arrhenius: Acids produce $$\chem{H^+}$$ ions in solution, bases produce $$\chem{OH^-}$$ ions.
• Brønsted-Lowry: Acids are proton ($$\chem{H^+}$$) donors, bases are proton acceptors.
$$\chem{\underbrace{HCl}_{acid} + \underbrace{H_2O}_{base} \rightleftharpoons Cl^- + H_3O^+}$$

Acid in Water

$$\chem{\underbrace{HA(aq)}_{Acid} + \underbrace{H_2O(l)}_{Base} \rightleftharpoons \underbrace{H_3O^+(aq)}_{\text{Conjugate base}} + \underbrace{A^-(aq)}_{\text{Conjugate base}}}$$

• Conjugate base is everything that remains of the acid molecule after a proton is lost.
• Conjugate acid is formed when the proton is transferred to the base.

Acid Strength

• Strong acid:
• Ionization equilibrium lies far to the right.
• Yields a weak conjugate base.
• Weak acid:
• Ionization equilibrium lies far to the left.
• Weaker the acid, stronger its conjugate base.

Various Ways to Describe Acid Strength

Property Strong Acid Weak Acid
$$K_a$$ value $$K_a$$ is large $$K_a$$ is small
Position of the dissociation (ionization) equilibrium Far to the right Far to the left
Equilibrium concentration of $$\conc{H^+}$$ compared with original $$\conc{H^+}\approx \conc{HA}_0$$ $$\conc{H^+}\ll \conc{HA}_0$$
Strength of conjugate base compared with that of water $$\chem{A^-}$$ much weakber base than $$\chem{H_2O}$$ $$\chem{A^-}$$ much stronger base than $$\chem{H_2O}$$

Water as an Acid and a Base

• Water is amphoteric:
• Behaves either as an acid or as a base.
• At $$25^\circ\mathrm{C}$$: $$K_w = \conc{H^+}\conc{OH^-} = 1.0 \times 10^{–14}$$
• No matter what the solution contains, the product of $$\conc{H^+}$$ and $$\conc{OH^-}$$ must always equal $$1.0 \times 10^{-14}$$ at $$25^\circ\mathrm{C}$$.
• Three possible situations:
• $$\conc{H^+}=\conc{OH^-}$$: neutral solution
• $$\conc{H^+}>\conc{OH^-}$$: acidic solution
• $$\conc{OH^-}>\conc{H^+}$$: basic solution

The pH Scale

• $$pH = –\log\conc{H^+}$$
• pH changes by 1 for every power of 10 change in $$\conc{H^+}$$.
• A compact way to represent solution acidity.
• pH decreases as $$\conc{H^+}$$ increases.
• Significant figures:
• The number of decimal places in the log is equal to the number of significant figures in the original number.
• pH Range
• $$pH = 7$$: neutral
• $$pH > 7$$: basic
• $$pH < 7$$: acidic

Exercise

Calculate the pH for each of the following solutions.

1. $$1.0 \times 10^{-4}\,\mathrm{M}\,\chem{H^+}$$
2. $$pH=4.00$$

3. $$0.0400\,\mathrm{M}\,\chem{OH^-}$$
4. $$pH=12.60$$

pH and pOH

• Recall:
$$K_w=\conc{H^+}\conc{OH^-}$$ $$-\log K_w = -\log \conc{H^+} - \log \conc{OH^-}$$ $$\mathrm{p}K_w=\mathrm{pH}+\mathrm{pOH}$$ $$14.00 = \mathrm{pH} + \mathrm{pOH}$$

Exercise

Calculate the pOH for each of the following solutions.

1. $$1.0 \times 10^{-4}\,\mathrm{M}\,\chem{H^+}$$
2. $$\mathrm{pOH}=10.00$$

3. $$0.040\,\mathrm{M}\,\chem{OH^-}$$
4. $$\mathrm{pOH}=1.40$$

Exercise

The pH of a solution is 5.85. What is the [OH–] for this solution?

$$\conc{OH^-}=7.1\times 10^{-9}\,\mathrm{M}$$

• What are the major species in solution?
• What is the dominant reaction that will take place?
• Is it an equilibrium reaction or a reaction that will go essentially to completion?
• React all major species until you are left with an equilibrium reaction.
• Solve for the pH if needed.

Concept Check

Calculate the pH of a $$1.5 \times 10^{–11}\,\mathrm{M}$$ solution of $$\chem{HCl}$$.

$$\mathrm{pH}=7.00$$

Concept Check

Calculate the pH of a $$1.5 \times 10^{–2}\,\mathrm{M}$$ solution of $$\chem{HNO_3}$$.

$$\mathrm{pH}=1.82$$

Solving Weak Acid Equilibrium Problems

1. List the major species in the solution.
2. Choose the species that can produce $$\chem{H^+}$$, and write balanced equations for the reactions producing $$\chem{H^+}$$.
3. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing $$\chem{H^+}$$.
4. Write the equilibrium expression for the dominant equilibrium.
5. List the initial concentrations of the species participating in the dominant equilibrium.
6. Define the change needed to achieve equilibrium; that is, define $$x$$.
7. Write the equilibrium concentrations in terms of $$x$$.
8. Substitute the equilibrium concentrations into the equilibrium expression.

Solving Weak Acid Equilibrium Problems (cont.)

1. Solve for $$x$$ the “easy” way, that is, by assuming that $$\conc{HA}_0-x$$ about equals $$\conc{HA}_0$$.
2. Use the 5% rule to verify whether the approximation is valid.
3. Calculate $$\conc{H^+}$$ and $$\mathrm{pH}$$.

Consider This

\begin{align} \chem{HCN(aq)+H_2O(l)} &\rightleftharpoons \chem{H_3O^+(aq)+CN^-(aq)} & K_a&=6.2 \cdot 10^{-1} \\ \chem{H_2O(l)+H_2O(l)} &\rightleftharpoons \chem{H_3O^+(aq)+OH^-(aq)} & K_w&=1.0 \cdot 10^{-14} \end{align} Which reaction controls the pH? Explain.

Exercise

Calculate the pH of a $$0.50\,\mathrm{M}$$ aqueous solution of the weak acid $$\chem{HF}$$. ($$K_a=7.2\times 10^{-4}$$)

\begin{align} & \chem{HF(aq)} &+& \chem{H_2O} &\rightleftharpoons& \chem{H_3O^+(aq)} &+& \chem{F^-(aq)} \\ \mathrm{Initial}\;\; & 0.50\,\mathrm{M} && - && \sim 0 && \sim 0 \\ \mathrm{Change}\;\; & -x && - && +x && +x \\ \mathrm{Equilibrium}\;\; & 0.50-x && - && x && x \\ \end{align} $$\mathrm{pH}=1.72$$

Percent Dissociation (Ionization)

$$\text{Percent dissociation}=\frac{\text{amount dissociated}\,\left(\bfrac{\mathrm{mol}}{\mathrm{L}}\right)}{\text{initial concentration}\,\left(\bfrac{\mathrm{mol}}{\mathrm{L}}\right)} \times 100%$$
• For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

Exercise

A solution of $$8.00\,\mathrm{M}$$ formic acid ($$\chem{HCHO_2}$$) is 0.47% ionized in water.

Calculate the $$K_a$$ value for formic acid.

$$K_a=1.8 \times 10^{-4}$$

Exercise

Calculate the pH of an $$8.00\,\mathrm{M}$$ solution of formic acid. Use the data from the previous slide to help you solve this problem.

$$\mathrm{pH}=1.42$$

Bases

• Arrhenius: bases produce $$\chem{OH^–}$$ ions.
• Brønsted–Lowry: bases are proton acceptors.
• In a basic solution at $$25^\circ \mathrm{C}$$, $$\mathrm{pH}>7.$$
• Ionic compounds containing $$\chem{OH^-}$$ are generally considered strong bases.
• $$\chem{LiOH}$$, $$\chem{NaOH}$$, $$\chem{KOH}$$, $$\chem{Ca(OH)_2}$$
• $$\mathrm{pOH} = –\log\conc{OH^-}$$
• $$\mathrm{pH} = 14.00 – \mathrm{pOH}$$

Concept Check

Calculate the $$\mathrm{pH}$$ of a $$1.0 \times 10^{–3}\,\mathrm{M}$$ solution of sodium hydroxide.

$$\mathrm{pH}=11.00$$

Concept Check

Calculate the $$\mathrm{pH}$$ of a $$1.0 \times 10^{–3}\,\mathrm{M}$$ solution of calcium hydroxide.

$$\mathrm{pH}=11.30$$

Weak Bases

• Equilibrium expression for weak bases uses $$K_n$$.
• $$\chem{CN^-(aq)+H_2O(l) \rightleftharpoons HCN(aq)+OH^-(aq)}$$ $$K_b=\frac{\conc{HCN}\conc{OH^-}}{\conc{CN^-}}$$
• $$\mathrm{pH}$$ calculations for solutions of weak bases are very similar to those for weak acids.
• $$K_w = \conc{H^+}\conc{OH^-} = 1.0 \times 10^{–14}$$
• $$\mathrm{pOH} = –\log\conc{OH^-}$$
• $$\mathrm{pH} = 14.00 – \mathrm{pOH}$$

Concept Check

Calculate the $$\mathrm{pH}$$ of a $$2.0\,\mathrm{M}$$ solution of ammonia ($$\chem{NH_3}$$). $$(K_b=1.8 \times 10^{-5})$$

$$\mathrm{pH}=11.78$$

Polyprotic Acids

• Acids that can furnish more than one proton.
• Always dissociates in a stepwise manner, one proton at a time.
• The conjugate base of the first dissociation equilibrium becomes the acid in the second step.
• For a typical weak polyprotic acid: $$K_{a1} > K_{a2} > K_{a3}$$
• For a typical polyprotic acid in water, only the first dissociation step is important to $$\mathrm{pH}$$.

Exercise

Calculate the $$\mathrm{pH}$$ of a $$1.00\,\mathrm{M}$$ solution of $$\chem{H_3PO_4}$$. \begin{align} K_{a1} &= 7.5 \times 10^{-3} \\ K_{a2} &= 6.2 \times 10^{-8} \\ K_{a3} &= 4.8 \times 10^{-13} \end{align}

$$\mathrm{pH}=1.08$$

Concept Check

Calculate the equilibrium concentration of $$\chem{PO_4^{3-}}$$ in a $$1.00\,\mathrm{M}$$ solution of $$\chem{H_3PO_4}$$. \begin{align} K_{a1} &= 7.5 \times 10^{-3} \\ K_{a2} &= 6.2 \times 10^{-8} \\ K_{a3} &= 4.8 \times 10^{-13} \end{align}

$$\conc{PO_4^{3-}}=3.6 \times 10^{-19}\,\mathrm{M}$$

Salts

• Ionic compounds.
• When dissolved in water, break up into its ions (which can behave as acids or bases).
• The salt of a strong acid and a strong base gives a neutral solution.
• $$\chem{KCl}$$, $$\chem{NaNO_3}$$
• A basic solution is formed if the anion of the salt is the conjugate base of a weak acid.
• $$\chem{NaF}$$, $$\chem{KC_2H_3O_2}$$
• $$K_w = K_a \times K_b$$
• Use $$K_b$$ when starting with base.

More of Salts

• An acidic solution is formed if the cation of the salt is the conjugate acid of a weak base.
• $$\chem{NH_4Cl}$$
• $$K_w = K_a \times K_b$$
• Use $$K_a$$ when starting with acid.

Summary of Properties of Salts

Cation Anion Acidic or Basic Example
neutral neutral neutral $$\chem{NaCl}$$
neutral conjugate base of weak acid basic $$\chem{NaF}$$
conjugate acid of a weak base neutral acidic $$\chem{NH_4Cl}$$
conjugate acid of a weak base conjugate acid of a weak acid depends on $$K_a$$ & $$K_b$$ values $$\chem{Al_2(SO_4)_3}$$

Qualitative Prediction of pH of Salt Solutions (from Weak Parents)

 $$K_a>K_b$$ $$\mathrm{pH}<7$$ (acidic) $$K_b>K_a$$ $$\mathrm{pH}>7$$ (basic) $$K_a=K_b$$ $$\mathrm{pH}=7$$ (neutral)

Exercise

Calculate the $$\mathrm{pH}$$ of a $$0.75\,\mathrm{M}$$ aqueous solution of $$\chem{NaCN}$$.$$K_a$$ for $$\chem{HCN}$$ is $$6.2 \times 10^{–10}$$.

\begin{align} & \chem{CN^-(aq)} &+& \chem{H_2O} &\rightleftharpoons& \chem{HCN(aq)} &+& \chem{OH^-(aq)} \\ \mathrm{Initial}\;\; & 0.75\,\mathrm{M} && - && 0 && \sim 0 \\ \mathrm{Change}\;\; & -x && - && +x && +x \\ \mathrm{Equilibrium}\;\; & 0.75-x && - && x && x \\ \end{align} $$\mathrm{pH}=11.54$$

Models of Acids and Bases

• Two factors for acidity in binary compounds:
• Bond Polarity (high is good)
• Bond Strength (low is good)
H-X Bond Bond Strength (kJ/mol) Acid Strength in Water
H-F 565 Weak
H-Cl 427 Strong
H-Br 363 Strong
H-I 295 Strong

Oxyacids

• Contains the group H–O–X.
• For a given series the acid strength increases with an increase in the number of oxygen atoms attached to the central atom.
• The greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule.

Comparison of Electronegativitiy of X and $$K_a$$ Value for a Series of Oxyacids

Acid X Electronegativity of X $$K_a$$ for Acid
$$\chem{HOCl}$$ $$\chem{Cl}$$ 3.0 $$4 \times 10^{-8}$$
$$\chem{HOBr}$$ $$\chem{Br}$$ 2.8 $$2 \times 10^{-9}$$
$$\chem{HOI}$$ $$\chem{I}$$ 2.5 $$2 \times 10^{-11}$$
$$\chem{HOCH_3}$$ $$\chem{CH_3}$$ 2.3 (for carbon $$\chem{CH_3}$$) $$\sim 10^{-15}$$

Oxides

• Acidic Oxides (Acid Anhydrides):
• O—X bond is strong and covalent. $$\chem{SO_2}, \chem{NO_2}, \chem{CO_2}$$
• When H—O—X grouping is dissolved in water, the O—X bond will remain intact. It will be the polar and relatively weak H—O bond that will tend to break, releasing a proton.
• Basic Oxides (Basic Anhydrides):
• O—X bond is ionic. $$\chem{K_2O}, \chem{CaO}$$
• If X has a very low electronegativity, the O—X bond will be ionic and subject to being broken in polar water, producing a basic solution.

Lewis Acids and Bases

• Lewis acid: electron pair acceptor
• Lewis base: electron pair donor

• In the above reaction, $$\chem{Al^{3+}}$$ is acting as a Lewis acid.
• In the above reaction, $$\chem{H_2O}$$ is acting as a Lewis base.

Three Models for Acids and Bases

Model Definition of Acid Definition of Base
Arrhenius $$\chem{H^+}$$ producer $$\chem{OH^-}$$ producer
Brønsted-Lowry $$\chem{H^+}$$ donor $$\chem{H^+}$$ acceptor
Lewis Electron-pair acceptor Electron-pair donor

When analyzing an acid-base equilibrium problem:

• Ask this question: What are the major species in the solution and what is their chemical behavior?
• What major species are present?
• Does a reaction occur that can be assumed to go to completion?
• What equilibrium dominates the solution?
• Let the problem guide you. Be patient.

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