AP Chemistry 8.1-8.2: Acid-Base Theory and pH

Brønsted–Lowry acid

A species that donates a proton, $H^+$, during an acid–base reaction.

Brønsted–Lowry base

A species that accepts a proton, $H^+$, during an acid–base reaction.

A solution has $[OH^-]=1.25\times10^{-2}\,\mathrm{M}$ at $25\,^{\circ}\mathrm{C}$. What are its pOH and pH?

$\mathrm{pOH}=-\log(1.25\times10^{-2})=1.903$. Therefore, $\mathrm{pH}=14.00-1.903=12.10$.

How does a Brønsted–Lowry acid–base reaction differ from an Arrhenius acid–base description?

The Brønsted–Lowry model defines acids and bases by proton transfer and applies beyond reactions that directly produce $H^+$ or $OH^-$ in water. The Arrhenius model specifically describes acids producing hydronium and bases producing hydroxide in aqueous solution.

Conjugate base

The species formed when a Brønsted–Lowry acid donates a proton. It has one fewer hydrogen and one less positive charge, or one more negative charge, than the original acid.

Conjugate acid

The species formed when a Brønsted–Lowry base accepts a proton. It has one more hydrogen and one more positive charge than the original base.

In the reaction $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$, identify the acid, base, and both conjugate partners.

$H_2O$ is the acid and $NH_3$ is the base. Their conjugates are $OH^-$, the conjugate base of water, and $NH_4^+$, the conjugate acid of ammonia.

How can conjugate acid–base pairs be recognized in a chemical equation?

The members of a conjugate pair differ by exactly one proton, $H^+$. The acid has one more proton than its conjugate base, while the conjugate acid has one more proton than its base.

What is acid ionization in water?

Acid ionization is the transfer of a proton from an acid to a water molecule. Water becomes $H_3O^+$, while the acid becomes its conjugate base; for example, $HF + H_2O \rightleftharpoons H_3O^+ + F^-$.

What is base ionization in water?

Base ionization occurs when a base accepts a proton from water. Water becomes $OH^-$, while the base becomes its conjugate acid; for example, $C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^-$.

Why is water amphiprotic?

Water can donate a proton, acting as an acid, or accept a proton, acting as a base. For example, it donates $H^+$ to $NH_3$ but accepts $H^+$ from $HF$.

Amphiprotic substance

A species that can both donate and accept a proton in Brønsted–Lowry reactions. Bicarbonate, $HCO_3^-$, is amphiprotic because it can become either $CO_3^{2-}$ or $H_2CO_3$.

Write the reaction showing bicarbonate acting as a Brønsted–Lowry acid in water.

$HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$. Bicarbonate donates a proton and water accepts it.

Write the reaction showing bicarbonate acting as a Brønsted–Lowry base in water.

$HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$. Bicarbonate accepts a proton from water.

How can $HSO_3^-$ act as both an acid and a base?

As an acid, it can donate a proton: $HSO_3^- + OH^- \rightleftharpoons SO_3^{2-} + H_2O$. As a base, it can accept a proton: $HSO_3^- + HI \rightleftharpoons H_2SO_3 + I^-$.

Autoionization

A process in which molecules of an amphiprotic substance react with one another to form ions. Water undergoes autoionization according to $2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$.

Ion-product constant of water, $K_w$

The equilibrium constant for water autoionization: $K_w = [H_3O^+][OH^-]$. At $25\,^{\circ}\mathrm{C}$, $K_w = 1.0\times10^{-14}$.

What are the hydronium and hydroxide concentrations in pure water at $25\,^{\circ}\mathrm{C}$?

They are equal because water autoionization produces the ions in a 1:1 ratio. Thus, $[H_3O^+] = [OH^-] = \sqrt{K_w} = 1.0\times10^{-7}\,\mathrm{M}$.

An aqueous solution has $[H_3O^+] = 2.0\times10^{-6}\,\mathrm{M}$ at $25\,^{\circ}\mathrm{C}$. What is $[OH^-]$?

Use $[OH^-] = K_w/[H_3O^+]$: $[OH^-] = (1.0\times10^{-14})/(2.0\times10^{-6}) = 5.0\times10^{-9}\,\mathrm{M}$.

How does adding an acid affect the water autoionization equilibrium?

Adding acid increases $[H_3O^+]$, so the equilibrium $2H_2O \rightleftharpoons H_3O^+ + OH^-$ shifts left by Le Châtelier’s principle. Consequently, $[OH^-]$ decreases so that $[H_3O^+][OH^-]$ remains equal to $K_w$ at that temperature.

How does temperature affect $K_w$ for water?

Water autoionization is endothermic, so increasing temperature increases $K_w$ and the equilibrium concentrations of both $H_3O^+$ and $OH^-$. At $80\,^{\circ}\mathrm{C}$, for example, $K_w=2.4\times10^{-13}$.

How are aqueous solutions classified as acidic, neutral, or basic using ion concentrations?

An acidic solution has $[H_3O^+] > [OH^-]$, a neutral solution has $[H_3O^+] = [OH^-]$, and a basic solution has $[H_3O^+] < [OH^-]$.

p-function

A logarithmic notation defined by $pX=-\log X$. It allows very large or small quantities, such as ion concentrations, to be expressed conveniently.

pH

The negative base-10 logarithm of the hydronium ion concentration: $\mathrm{pH}=-\log[H_3O^+]$. Equivalently, $[H_3O^+]=10^{-\mathrm{pH}}$.

pOH

The negative base-10 logarithm of the hydroxide ion concentration: $\mathrm{pOH}=-\log[OH^-]$. Equivalently, $[OH^-]=10^{-\mathrm{pOH}}$.

What relationship connects pH, pOH, and $K_w$?

Taking negative logarithms of $K_w=[H_3O^+][OH^-]$ gives $pK_w=\mathrm{pH} + \mathrm{pOH}$. At $25\,^{\circ}\mathrm{C}$, $pK_w=14.00$, so $\mathrm{pH} + \mathrm{pOH}=14.00$.

What are the pH and pOH criteria for acidic, neutral, and basic solutions at $25\,^{\circ}\mathrm{C}$?

Acidic solutions have $\mathrm{pH}<7.00$ and $\mathrm{pOH}>7.00$; neutral solutions have $\mathrm{pH}=\mathrm{pOH}=7.00$; basic solutions have $\mathrm{pH}>7.00$ and $\mathrm{pOH}<7.00$.

Why is a neutral solution not always pH 7?

Neutrality means $[H_3O^+]=[OH^-]$, not necessarily that each concentration equals $1.0\times10^{-7}\,\mathrm{M}$. Because $K_w$ changes with temperature, the neutral pH changes; at $80\,^{\circ}\mathrm{C}$, neutral water has pH about $6.31$.

Calculate the pH of a solution with $[H_3O^+]=1.2\times10^{-3}\,\mathrm{M}$.

$\mathrm{pH}=-\log(1.2\times10^{-3})=2.92$. The solution is acidic.

Calculate $[H_3O^+]$ when a solution has pH $=7.3$.

Use $[H_3O^+]=10^{-\mathrm{pH}}$: $[H_3O^+]=10^{-7.3}\approx5.0\times10^{-8}\,\mathrm{M}$.

How does a strong acid behave in aqueous solution?

A strong acid ionizes essentially completely in water. For a monoprotic strong acid such as HCl, the acid concentration approximately equals the hydronium concentration: $[H_3O^+]\approx[HCl]$.

How does a strong base behave in aqueous solution?

A strong base dissociates essentially completely in water. For a soluble metal hydroxide such as NaOH, $[OH^-]\approx[NaOH]$; for a compound such as $Ba(OH)_2$, each formula unit produces two hydroxide ions, so $[OH^-]\approx2[Ba(OH)_2]$.

What is the relationship between the concentration of a strong acid and its pH?

For a monoprotic strong acid at ordinary concentrations, complete ionization gives $[H_3O^+]$ approximately equal to the acid's molar concentration. Then calculate $\mathrm{pH}=-\log[H_3O^+]$.

What is the relationship between the concentration of a strong base and its pOH?

For a strong base that produces one hydroxide ion per formula unit, complete dissociation gives $[OH^-]$ approximately equal to the base's molar concentration. Then calculate $\mathrm{pOH}=-\log[OH^-]$ and use $\mathrm{pH}=14.00-\mathrm{pOH}$ at $25\,^{\circ}\mathrm{C}$.

Calculate the pH of $0.010\,\mathrm{M}$ HCl at $25\,^{\circ}\mathrm{C}$.

HCl is a strong monoprotic acid, so $[H_3O^+]=0.010\,\mathrm{M}$. Therefore, $\mathrm{pH}=-\log(1.0\times10^{-2})=2.00$.

Calculate the pH of $0.020\,\mathrm{M}$ $Ba(OH)_2$ at $25\,^{\circ}\mathrm{C}$.

$Ba(OH)_2$ dissociates completely and produces two hydroxide ions per formula unit, so $[OH^-]=2(0.020)=0.040\,\mathrm{M}$. Thus, $\mathrm{pOH}=-\log(0.040)=1.40$ and $\mathrm{pH}=14.00-1.40=12.60$.

How does a tenfold change in hydronium concentration affect pH?

Because pH is logarithmic, increasing $[H_3O^+]$ by a factor of 10 lowers pH by 1 unit. Decreasing $[H_3O^+]$ by a factor of 10 raises pH by 1 unit.

How should significant figures be handled in pH calculations?

The number of digits after the decimal point in a logarithmic quantity, such as pH or pOH, should equal the number of significant figures in the original concentration. For example, $[H_3O^+]=1.0\times10^{-3}\,\mathrm{M}$ gives $\mathrm{pH}=3.00$.

Can pH be less than 0 or greater than 14?

Yes. Although many introductory diagrams show a pH range of 0 to 14, very concentrated strong acids can have pH below 0 and very concentrated bases can have pH above 14.