The concentrations of hydronium ion ([H3O+]) and hydroxide ion ([OH–]) are reported as pH and pOH, respectively.
pH formula: pH = –log[H3O+]
pOH formula: pOH = –log[OH–]
There is an interchangeability between the terms "hydrogen ion" and "hydronium ion." While H+(aq) is commonly used, H3O+(aq) is preferred, particularly in the context of the AP Exam.
Water autoionizes with an equilibrium constant represented as Kw:
Kw = [H3O+][OH–] = 1.0 × 10–14 at 25°C.
In pure water, it is noted that pH = pOH, defining pure water as a neutral solution. At 25°C, pKw is 14.0, leading to pH and pOH both equating to 7.0.
Relationship: pKw = 14 = pH + pOH at 25°C.
The value of Kw varies with temperature, indicating that the pH of neutral pure water will not always equal 7.0 outside of 25°C.
A Brønsted-Lowry acid is defined as a proton donor, while a Brønsted-Lowry base serves as a proton acceptor.
Water plays a crucial role in many acid-base reactions due to its molecular structure, allowing it to both donate and accept protons from dissolved species.
Reaction: HCN(aq) + H2O(l) ⇌ CN–(aq) + H3O+(aq)
Acid (Reactant): HCN
Conjugate Base (Product): CN–
Base (Reactant): H2O
Conjugate Acid (Product): H3O+
Reaction: CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH–(aq)
Acid (Reactant): CH3NH3+
Conjugate Base (Product): CH3NH2
Base (Reactant): H2O
Conjugate Acid (Product): H3O+
In pure water, a molecule of H2O functions as an acid, donating a H+ ion, while another accepts it, portraying the dynamics of autoionization.
Autoionization reaction: 2 H2O(l) ⇌ H3O+(aq) + OH–(aq)
The equilibrium constant Kw remains consistent at 25°C: Kw = [H3O+][OH–] = 1.0 × 10–14.
The letter "p" in formulas (pH, pOH, pKw) implies taking the negative logarithm of a concentration.
Concentrations of H+ or OH– have units of mol/L (M), while pH and pOH are dimensionless.
To calculate values using [H3O+] and [OH–]:
If [H3O+] is known, pH can be calculated as: pH = –log[H3O+]
To find [H3O+] from pH: [H3O+] = 10–pH
If [OH–] is known, pOH can be calculated using: pOH = –log[OH–]
And to find [OH–] from pOH: [OH–] = 10–pOH
Kw = [H3O+][OH–] = 1.0 × 10–14
pKw = pH + pOH = 14
The significance of digits in pH is determined by the number of significant figures in [H3O+]. The following examples illustrate how changes in concentration affect pH accuracy:
[H3O+] = 1 × 10–3 M → pH = 3.0
[H3O+] = 1.2 × 10–3 M → pH = 2.92
[H3O+] = 1.22 × 10–3 M → pH = 2.914
Students need to fill in tables for [H3O+], pH, [OH–], pOH while considering significant digits. Elements in the table also help reinforce the concept that:
As [H3O+] increases, pH decreases.
As [OH–] increases, pH increases.
The reaction is endothermic, where the equilibrium constant Kw changes with temperature.
The following values encapsulate changes in Kw with temperature increases:
Temperature (°C) | Kw | pKw | pH | pOH
0°C → 1.14 × 10–15; pKw = 14.943
10°C → 2.93 × 10–15; pKw = 14.533
20°C → 6.81 × 10–15; pKw = 14.167
25°C → 1.01 × 10–14; pKw = 13.996
30°C → 1.47 × 10–14; pKw = 13.833
40°C → 2.92 × 10–14; pKw = 13.535
50°C → 5.48 × 10–14; pKw = 13.261
Analyze statements regarding pure H2O:
A sample of pure H2O cannot have a pH of 7.0 at all temperatures.
[H+] equals [OH–] in pure H2O at all temperatures.
A sample of pure H2O is defined as neutral if pH = pOH.
If the pH is less than 7, the temperature must be less than 25°C.
H2O ionizes more at 50°C than at 25°C.
As temperature rises, pH of pure H2O declines.
If the pH of pure H2O is 6.8, pOH will be 7.2.