Acid/base equilibrium is a reversible reaction between a Brønsted acid (a proton donor) and a Brønsted base (a proton acceptor). When the acid donates a proton, it becomes its conjugate base. When the base accepts a proton, it becomes its conjugate acid.

Ionization of water

A classic example is the ionization of water:

$H_{2} O \leftrightarrow H^{+} O H^{-}$

The product of the hydrogen ion concentration ( $[H^{+}]$ ) and the hydroxide ion concentration ( $[O H^{-}]$ ) is the ion-product constant for water, $K_{w}$ , which is approximately $1 \times 1 0^{- 14}$ at $25°$ C.

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration:

$p H = - l o g [H^{+}]$

For pure water, $[H^{+}] = 1 0^{- 7} M$ , so the pH is $7$ , which indicates a neutral solution.

A solution is acidic if its pH is less than $7$ and basic if its pH is greater than $7$ . The pOH is defined as $- l o g [O H^{-}]$ , and at $25°$ C:

$pH + pOH = 14$ .

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Acid/base equilibria describe how acids and bases interconvert with their conjugate acids and conjugate bases. The key idea is simple:

When an acid donates a proton, it forms its conjugate base.
When a base accepts a proton, it forms its conjugate acid.

Inverse proportional relationship between Ka and Kb for conjugate acid-base pairs

In strong acids and strong bases, complete dissociation occurs because their conjugate species are highly stable.
Weak acids and weak bases only partially dissociate due to the moderate stability of their conjugate species.
This equilibrium can be shifted by the presence of a common ion, as explained by Le Chatelier’s principle; for instance, acetic acid ( $C H_{3} OO H$ ) dissociates less in the presence of its salt, $C H_{3} COON a$ , and ammonium hydroxide ( $N H_{4} O H$ ) shows reduced dissociation when $N H_{4} Cl$ is added.

Hydrolysis of salts of weak acids or bases

Example of a salt of a weak acid:
- Sodium acetate ( $C H_{3} COON a$ )**, derived from acetic acid ( $C H_{3} COO H$ ) - a weak acid.
Example of a salt of a weak base:
- Ammonium chloride ( $N H_{4} Cl$ )**, formed from ammonia ( $N H_{3}$ ) - a weak base.

Calculation of pH of solutions of salts of weak acids or bases

Consider a 0.10 M aqueous solution of sodium acetate ( $C H_{3} COON a$ ). In water, sodium acetate dissociates completely into $N a^{+}$ and $C H_{3} CO O^{-}$ (the acetate ion). The relevant equilibrium is the hydrolysis of the acetate ion:

$CH_{3} COO^{-} + H_{2} O ⇌ CH_{3} COOH + O H^{-}$

Acid/base constants:

Acetic acid ( $C H_{3} COO H$ ) has a dissociation constant of $K a = 1.8 \times 1 0^{- 5}$ . Because $C H_{3} CO O^{-}$ is the conjugate base of a weak acid, its basicity is described by $K_{b}$ , which we find using:

$K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}} \approx 5.6 \times 1 0^{- 10} .$

Setting up the equilibrium expression:

Let $x$ be the concentration of $O H^{-}$ produced by acetate hydrolysis. The equilibrium expression for the reaction is:

$K_{b} = \frac{[ CH _{3} COOH ] [ O H ^{-} ]}{[ CH _{3} COO ^{-} ]} .$

Initially,

$[CH_{3} COO^{-}] = 0.10 M, [O H^{-}] = 0.$

At equilibrium,

$[CH_{3} COOH] \approx x, [O H^{-}] = x, and [CH_{3} COO^{-}] \approx 0.10 - x .$

Because $x$ is very small, we approximate:

$[CH_{3} COO^{-}] \approx 0.10 M .$

Substituting into the equilibrium expression gives:

$5.6 \times 1 0^{- 10} = \frac{x ^{2}}{0.10} .$

Solving for $x$ :

$x^{2} = (5.6 \times 1 0^{- 10}) \times 0.10 = 5.6 \times 1 0^{- 11},$

$x = 5.6 \times 1 0^{- 11} \approx 7.5 \times 1 0^{- 6} .$

Therefore,

$[O H^{-}] = 7.5 \times 1 0^{- 6} M .$

Calculating pOH and pH:

$pOH = - lo g [O H^{-}] = - lo g (7.5 \times 1 0^{- 6}) \approx 5.12,$

$pH = 14 - pOH = 14 - 5.12 = 8.88.$

Hence, a $0.10 M$ solution of sodium acetate is basic, with a $p H \approx 8.88$ . This happens because the acetate anion acts as a weak base and partially hydrolyzes in water, producing $O H^{-}$ and raising the pH above $7$ .

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

The acid dissociation constant ( $K_{a}$ ) describes how much an acid donates protons ( $H^{+}$ ) in water. A general form is:

$H-Acid \leftrightarrow H^{+} + Acid^{-}$

Similarly, the base dissociation constant ( $K_{b}$ ) describes how strongly a base produces hydroxide by accepting a proton from water:

$Base + H_{2} O \leftrightarrow H-Base^{+} + O H^{-}$

Note that water is not included in these expressions because its concentration remains constant.
For a conjugate acid-base pair, the product of $K_{a}$ and $K_{b}$ equals the ionization constant of water ( $K_{w}$ ), typically $1 0^{- 14}$ . To make these constants easier to compare, we define pKa and pKb as the negative logarithms of $K_{a}$ and $K_{b}$ , respectively:

$pKa = - lo g K_{a} and pKb = - lo g K_{b}$

and for any conjugate pair,

$pKa + pKb = 14$

Buffers

Buffers are solutions that resist changes in pH by maintaining an equilibrium between an acidic species and its conjugate base. In a buffer:

the acidic component donates protons when the pH rises
the basic component accepts protons when the pH falls

Buffers are typically composed of salts derived from weak acids and weak bases. Maximum buffering capacity occurs when the concentrations of the acid and its conjugate base are equal. In that case, the pH equals the $p K_{a}$ of the acid (or, for a weak base, the pH equals $14$ minus the $p K_{b}$ ).

Influence on titration curves

On a titration curve, the buffering region appears as a relatively flat segment near the point of inflection. It typically spans about one pH unit above and below the $p K_{a}$ (or $14 - p K_{b}$ ).

Ionization of water

Water self-ionizes: $H_{2} O \leftrightarrow H^{+} + O H^{-}$
$K_{w} = [H^{+}] [O H^{-}] = 1 \times 1 0^{- 14}$ at $2 5^{\circ}$ C
pH = $- lo g [H^{+}]$ ; neutral pH = 7; pH + pOH = 14

Conjugate acids and bases

Acid donates proton → forms conjugate base
Base accepts proton → forms conjugate acid
Strong acids/bases: complete dissociation; weak acids/bases: partial dissociation

Hydrolysis of salts of weak acids or bases

Salt of weak acid (e.g., $C H_{3} COON a$ ): anion hydrolyzes, solution becomes basic
Salt of weak base (e.g., $N H_{4} Cl$ ): cation hydrolyzes, solution becomes acidic

Calculation of pH for salts of weak acids/bases

Use $K_{b} = K_{w} / K_{a}$ for conjugate base of weak acid
Set up equilibrium: $K_{b} = \frac{[ C H _{3} COO H ] [ O H ^{-} ]}{[ C H _{3} CO O ^{-} ]}$
Approximate $[O H^{-}]$ using $x^{2} = K_{b} \times [salt]$ ; pOH = $- lo g [O H^{-}]$ ; pH = $14 -$ pOH

Equilibrium constants: $K_{a}$ , $K_{b}$ , $p K_{a}$ , $p K_{b}$

$K_{a}$ : acid dissociation constant; $K_{b}$ : base dissociation constant
$K_{a} \times K_{b} = K_{w} = 1 0^{- 14}$
$p K_{a} = - lo g K_{a}$ ; $p K_{b} = - lo g K_{b}$ ; $p K_{a} + p K_{b} = 14$

Buffers

Resist pH changes; contain weak acid + conjugate base (or weak base + conjugate acid)
Maximum buffering: $[acid] = [conjugate base]$ , pH = $p K_{a}$
Buffer region: ±1 pH unit around $p K_{a}$ (or $14 - p K_{b}$ ) on titration curve

Acid/base equilibria

Ionization of water

A classic example is the ionization of water:

$H_{2} O \leftrightarrow H^{+} O H^{-}$

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration:

$p H = - l o g [H^{+}]$

For pure water, $[H^{+}] = 1 0^{- 7} M$ , so the pH is $7$ , which indicates a neutral solution.

A solution is acidic if its pH is less than $7$ and basic if its pH is greater than $7$ . The pOH is defined as $- l o g [O H^{-}]$ , and at $25°$ C:

$pH + pOH = 14$ .

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Acid/base equilibria describe how acids and bases interconvert with their conjugate acids and conjugate bases. The key idea is simple:

When an acid donates a proton, it forms its conjugate base.
When a base accepts a proton, it forms its conjugate acid.

In strong acids and strong bases, complete dissociation occurs because their conjugate species are highly stable.
Weak acids and weak bases only partially dissociate due to the moderate stability of their conjugate species.
This equilibrium can be shifted by the presence of a common ion, as explained by Le Chatelier’s principle; for instance, acetic acid ( $C H_{3} OO H$ ) dissociates less in the presence of its salt, $C H_{3} COON a$ , and ammonium hydroxide ( $N H_{4} O H$ ) shows reduced dissociation when $N H_{4} Cl$ is added.

Hydrolysis of salts of weak acids or bases

Example of a salt of a weak acid:
- Sodium acetate ( $C H_{3} COON a$ )**, derived from acetic acid ( $C H_{3} COO H$ ) - a weak acid.
Example of a salt of a weak base:
- Ammonium chloride ( $N H_{4} Cl$ )**, formed from ammonia ( $N H_{3}$ ) - a weak base.

Calculation of pH of solutions of salts of weak acids or bases

$CH_{3} COO^{-} + H_{2} O ⇌ CH_{3} COOH + O H^{-}$

Acid/base constants:

$K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}} \approx 5.6 \times 1 0^{- 10} .$

Setting up the equilibrium expression:

Let $x$ be the concentration of $O H^{-}$ produced by acetate hydrolysis. The equilibrium expression for the reaction is:

$K_{b} = \frac{[ CH _{3} COOH ] [ O H ^{-} ]}{[ CH _{3} COO ^{-} ]} .$

Initially,

$[CH_{3} COO^{-}] = 0.10 M, [O H^{-}] = 0.$

At equilibrium,

$[CH_{3} COOH] \approx x, [O H^{-}] = x, and [CH_{3} COO^{-}] \approx 0.10 - x .$

Because $x$ is very small, we approximate:

$[CH_{3} COO^{-}] \approx 0.10 M .$

Substituting into the equilibrium expression gives:

$5.6 \times 1 0^{- 10} = \frac{x ^{2}}{0.10} .$

Solving for $x$ :

$x^{2} = (5.6 \times 1 0^{- 10}) \times 0.10 = 5.6 \times 1 0^{- 11},$

$x = 5.6 \times 1 0^{- 11} \approx 7.5 \times 1 0^{- 6} .$

Therefore,

$[O H^{-}] = 7.5 \times 1 0^{- 6} M .$

Calculating pOH and pH:

$pOH = - lo g [O H^{-}] = - lo g (7.5 \times 1 0^{- 6}) \approx 5.12,$

$pH = 14 - pOH = 14 - 5.12 = 8.88.$

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

The acid dissociation constant ( $K_{a}$ ) describes how much an acid donates protons ( $H^{+}$ ) in water. A general form is:

$H-Acid \leftrightarrow H^{+} + Acid^{-}$

Similarly, the base dissociation constant ( $K_{b}$ ) describes how strongly a base produces hydroxide by accepting a proton from water:

$Base + H_{2} O \leftrightarrow H-Base^{+} + O H^{-}$

$pKa = - lo g K_{a} and pKb = - lo g K_{b}$

and for any conjugate pair,

$pKa + pKb = 14$

Buffers

Buffers are solutions that resist changes in pH by maintaining an equilibrium between an acidic species and its conjugate base. In a buffer:

the acidic component donates protons when the pH rises
the basic component accepts protons when the pH falls

Influence on titration curves

On a titration curve, the buffering region appears as a relatively flat segment near the point of inflection. It typically spans about one pH unit above and below the $p K_{a}$ (or $14 - p K_{b}$ ).

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4.6. Unique nature of water and its solutions

Acid/base equilibria

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Ionization of water

A classic example is the ionization of water:

$H_{2} O \leftrightarrow H^{+} O H^{-}$

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration:

$p H = - l o g [H^{+}]$

For pure water, $[H^{+}] = 1 0^{- 7} M$ , so the pH is $7$ , which indicates a neutral solution.

A solution is acidic if its pH is less than $7$ and basic if its pH is greater than $7$ . The pOH is defined as $- l o g [O H^{-}]$ , and at $25°$ C:

$pH + pOH = 14$ .

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Acid/base equilibria describe how acids and bases interconvert with their conjugate acids and conjugate bases. The key idea is simple:

When an acid donates a proton, it forms its conjugate base.
When a base accepts a proton, it forms its conjugate acid.

In strong acids and strong bases, complete dissociation occurs because their conjugate species are highly stable.
Weak acids and weak bases only partially dissociate due to the moderate stability of their conjugate species.
This equilibrium can be shifted by the presence of a common ion, as explained by Le Chatelier’s principle; for instance, acetic acid ( $C H_{3} OO H$ ) dissociates less in the presence of its salt, $C H_{3} COON a$ , and ammonium hydroxide ( $N H_{4} O H$ ) shows reduced dissociation when $N H_{4} Cl$ is added.

Hydrolysis of salts of weak acids or bases

Example of a salt of a weak acid:
- Sodium acetate ( $C H_{3} COON a$ )**, derived from acetic acid ( $C H_{3} COO H$ ) - a weak acid.
Example of a salt of a weak base:
- Ammonium chloride ( $N H_{4} Cl$ )**, formed from ammonia ( $N H_{3}$ ) - a weak base.

Calculation of pH of solutions of salts of weak acids or bases

$CH_{3} COO^{-} + H_{2} O ⇌ CH_{3} COOH + O H^{-}$

Acid/base constants:

$K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}} \approx 5.6 \times 1 0^{- 10} .$

Setting up the equilibrium expression:

Let $x$ be the concentration of $O H^{-}$ produced by acetate hydrolysis. The equilibrium expression for the reaction is:

$K_{b} = \frac{[ CH _{3} COOH ] [ O H ^{-} ]}{[ CH _{3} COO ^{-} ]} .$

Initially,

$[CH_{3} COO^{-}] = 0.10 M, [O H^{-}] = 0.$

At equilibrium,

$[CH_{3} COOH] \approx x, [O H^{-}] = x, and [CH_{3} COO^{-}] \approx 0.10 - x .$

Because $x$ is very small, we approximate:

$[CH_{3} COO^{-}] \approx 0.10 M .$

Substituting into the equilibrium expression gives:

$5.6 \times 1 0^{- 10} = \frac{x ^{2}}{0.10} .$

Solving for $x$ :

$x^{2} = (5.6 \times 1 0^{- 10}) \times 0.10 = 5.6 \times 1 0^{- 11},$

$x = 5.6 \times 1 0^{- 11} \approx 7.5 \times 1 0^{- 6} .$

Therefore,

$[O H^{-}] = 7.5 \times 1 0^{- 6} M .$

Calculating pOH and pH:

$pOH = - lo g [O H^{-}] = - lo g (7.5 \times 1 0^{- 6}) \approx 5.12,$

$pH = 14 - pOH = 14 - 5.12 = 8.88.$

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

The acid dissociation constant ( $K_{a}$ ) describes how much an acid donates protons ( $H^{+}$ ) in water. A general form is:

$H-Acid \leftrightarrow H^{+} + Acid^{-}$

Similarly, the base dissociation constant ( $K_{b}$ ) describes how strongly a base produces hydroxide by accepting a proton from water:

$Base + H_{2} O \leftrightarrow H-Base^{+} + O H^{-}$

$pKa = - lo g K_{a} and pKb = - lo g K_{b}$

and for any conjugate pair,

$pKa + pKb = 14$

Buffers

Buffers are solutions that resist changes in pH by maintaining an equilibrium between an acidic species and its conjugate base. In a buffer:

the acidic component donates protons when the pH rises
the basic component accepts protons when the pH falls

Influence on titration curves

On a titration curve, the buffering region appears as a relatively flat segment near the point of inflection. It typically spans about one pH unit above and below the $p K_{a}$ (or $14 - p K_{b}$ ).

Ionization of water

Water self-ionizes: $H_{2} O \leftrightarrow H^{+} + O H^{-}$
$K_{w} = [H^{+}] [O H^{-}] = 1 \times 1 0^{- 14}$ at $2 5^{\circ}$ C
pH = $- lo g [H^{+}]$ ; neutral pH = 7; pH + pOH = 14

Conjugate acids and bases

Acid donates proton → forms conjugate base
Base accepts proton → forms conjugate acid
Strong acids/bases: complete dissociation; weak acids/bases: partial dissociation

Hydrolysis of salts of weak acids or bases

Salt of weak acid (e.g., $C H_{3} COON a$ ): anion hydrolyzes, solution becomes basic
Salt of weak base (e.g., $N H_{4} Cl$ ): cation hydrolyzes, solution becomes acidic

Calculation of pH for salts of weak acids/bases

Use $K_{b} = K_{w} / K_{a}$ for conjugate base of weak acid
Set up equilibrium: $K_{b} = \frac{[ C H _{3} COO H ] [ O H ^{-} ]}{[ C H _{3} CO O ^{-} ]}$
Approximate $[O H^{-}]$ using $x^{2} = K_{b} \times [salt]$ ; pOH = $- lo g [O H^{-}]$ ; pH = $14 -$ pOH

Equilibrium constants: $K_{a}$ , $K_{b}$ , $p K_{a}$ , $p K_{b}$

$K_{a}$ : acid dissociation constant; $K_{b}$ : base dissociation constant
$K_{a} \times K_{b} = K_{w} = 1 0^{- 14}$
$p K_{a} = - lo g K_{a}$ ; $p K_{b} = - lo g K_{b}$ ; $p K_{a} + p K_{b} = 14$

Buffers

Resist pH changes; contain weak acid + conjugate base (or weak base + conjugate acid)
Maximum buffering: $[acid] = [conjugate base]$ , pH = $p K_{a}$
Buffer region: ±1 pH unit around $p K_{a}$ (or $14 - p K_{b}$ ) on titration curve

Acid/base equilibria

Ionization of water

A classic example is the ionization of water:

$H_{2} O \leftrightarrow H^{+} O H^{-}$

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration:

$p H = - l o g [H^{+}]$

For pure water, $[H^{+}] = 1 0^{- 7} M$ , so the pH is $7$ , which indicates a neutral solution.

A solution is acidic if its pH is less than $7$ and basic if its pH is greater than $7$ . The pOH is defined as $- l o g [O H^{-}]$ , and at $25°$ C:

$pH + pOH = 14$ .

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Acid/base equilibria describe how acids and bases interconvert with their conjugate acids and conjugate bases. The key idea is simple:

When an acid donates a proton, it forms its conjugate base.
When a base accepts a proton, it forms its conjugate acid.

In strong acids and strong bases, complete dissociation occurs because their conjugate species are highly stable.
Weak acids and weak bases only partially dissociate due to the moderate stability of their conjugate species.
This equilibrium can be shifted by the presence of a common ion, as explained by Le Chatelier’s principle; for instance, acetic acid ( $C H_{3} OO H$ ) dissociates less in the presence of its salt, $C H_{3} COON a$ , and ammonium hydroxide ( $N H_{4} O H$ ) shows reduced dissociation when $N H_{4} Cl$ is added.

Hydrolysis of salts of weak acids or bases

Example of a salt of a weak acid:
- Sodium acetate ( $C H_{3} COON a$ )**, derived from acetic acid ( $C H_{3} COO H$ ) - a weak acid.
Example of a salt of a weak base:
- Ammonium chloride ( $N H_{4} Cl$ )**, formed from ammonia ( $N H_{3}$ ) - a weak base.

Calculation of pH of solutions of salts of weak acids or bases

$CH_{3} COO^{-} + H_{2} O ⇌ CH_{3} COOH + O H^{-}$

Acid/base constants:

$K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}} \approx 5.6 \times 1 0^{- 10} .$

Setting up the equilibrium expression:

Let $x$ be the concentration of $O H^{-}$ produced by acetate hydrolysis. The equilibrium expression for the reaction is:

$K_{b} = \frac{[ CH _{3} COOH ] [ O H ^{-} ]}{[ CH _{3} COO ^{-} ]} .$

Initially,

$[CH_{3} COO^{-}] = 0.10 M, [O H^{-}] = 0.$

At equilibrium,

$[CH_{3} COOH] \approx x, [O H^{-}] = x, and [CH_{3} COO^{-}] \approx 0.10 - x .$

Because $x$ is very small, we approximate:

$[CH_{3} COO^{-}] \approx 0.10 M .$

Substituting into the equilibrium expression gives:

$5.6 \times 1 0^{- 10} = \frac{x ^{2}}{0.10} .$

Solving for $x$ :

$x^{2} = (5.6 \times 1 0^{- 10}) \times 0.10 = 5.6 \times 1 0^{- 11},$

$x = 5.6 \times 1 0^{- 11} \approx 7.5 \times 1 0^{- 6} .$

Therefore,

$[O H^{-}] = 7.5 \times 1 0^{- 6} M .$

Calculating pOH and pH:

$pOH = - lo g [O H^{-}] = - lo g (7.5 \times 1 0^{- 6}) \approx 5.12,$

$pH = 14 - pOH = 14 - 5.12 = 8.88.$

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

The acid dissociation constant ( $K_{a}$ ) describes how much an acid donates protons ( $H^{+}$ ) in water. A general form is:

$H-Acid \leftrightarrow H^{+} + Acid^{-}$

Similarly, the base dissociation constant ( $K_{b}$ ) describes how strongly a base produces hydroxide by accepting a proton from water:

$Base + H_{2} O \leftrightarrow H-Base^{+} + O H^{-}$

$pKa = - lo g K_{a} and pKb = - lo g K_{b}$

and for any conjugate pair,

$pKa + pKb = 14$

Buffers

Buffers are solutions that resist changes in pH by maintaining an equilibrium between an acidic species and its conjugate base. In a buffer:

the acidic component donates protons when the pH rises
the basic component accepts protons when the pH falls

Influence on titration curves

On a titration curve, the buffering region appears as a relatively flat segment near the point of inflection. It typically spans about one pH unit above and below the $p K_{a}$ (or $14 - p K_{b}$ ).

Acid/base equilibria

Ionization of water

Conjugate acids and bases (e.g., NH4+​ and NH3​)

Calculation of pH of solutions of salts of weak acids or bases

Acid/base constants:

Setting up the equilibrium expression:

Equilibrium constants Ka​ and Kb​: pKa​, pKb​

Buffers

Influence on titration curves

Acid/base equilibria

Ionization of water

Conjugate acids and bases (e.g., NH4+​ and NH3​)

Calculation of pH of solutions of salts of weak acids or bases

Acid/base constants:

Setting up the equilibrium expression:

Equilibrium constants Ka​ and Kb​: pKa​, pKb​

Buffers

Influence on titration curves

Acid/base equilibria

Ionization of water

Conjugate acids and bases (e.g., NH4+​ and NH3​)

Calculation of pH of solutions of salts of weak acids or bases

Acid/base constants:

Setting up the equilibrium expression:

Equilibrium constants Ka​ and Kb​: pKa​, pKb​

Buffers

Influence on titration curves

Acid/base equilibria

Ionization of water

Conjugate acids and bases (e.g., NH4+​ and NH3​)

Calculation of pH of solutions of salts of weak acids or bases

Acid/base constants:

Setting up the equilibrium expression:

Equilibrium constants Ka​ and Kb​: pKa​, pKb​

Buffers

Influence on titration curves

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$

Conjugate acids and bases (e.g., $N H_{4}^{+}$ and $N H_{3}$ )

Equilibrium constants $K_{a}$ and $K_{b}$ : $p K_{a}$ , $p K_{b}$