pH of Strong Bases

This section covers the calculation of pH for strong bases. Strong bases fully dissociate in water, releasing hydroxide ions $OH-$. The pH calculation depends on the number of hydroxide ions released by the base.

For monobasic compounds like NaOH, which release one OH- ion, the concentration of OH- equals the base concentration. The pOH is calculated first, then converted to pH:

Formula: pOH = -log$OH-$ = -log[base concentration] pH = 14 - pOH

For dibasic compounds like Ca(OH)2, which release two OH- ions, the concentration of OH- is twice the base concentration:

Formula: pOH = -log[2 × base concentration] pH = 14 - pOH

For polybasic compounds releasing n OH- ions, the general formula is:

Formula: pOH = -log[n × base concentration] pH = 14 - pOH

An example is provided for calculating the pH of 100 mL of 0.1 M NaOH:

Example: For 0.1 M NaOH, pOH = -log[0.1] = 1, pH = 14 - 1 = 13

This illustrates the application of pH calculations for strong bases.

pH of Water and Weak Acids

This section discusses the pH of pure water and weak acids. For water, the self-ionization equilibrium is considered, where the concentration of H3O+ equals OH-. The pH of pure water is calculated to be 7.

For weak acids, which partially dissociate in water, an equilibrium approach is used. The general equation for a weak acid HA is:

HA + H2O ⇌ H3O+ + A-

The equilibrium constant Ka is used to determine the concentration of H3O+. For weak acids, an approximation method is often used when the acid concentration is much larger than Ka.

Formula: $H3O+$ ≈ √(Ka × initial acid concentration)

An example is provided for calculating the pH of 500 mL of 0.250 M acetic acid (CH3COOH) with Ka = 1.75 × 10^-5:

Example: Using the quadratic equation or approximation method, $H3O+$ is calculated, and pH = -log$H3O+$

This demonstrates the more complex equilibrium considerations required for weak acid pH calculations.

pH of Weak Bases

This final section covers the calculation of pH for weak bases. Weak bases partially dissociate in water, establishing an equilibrium. The general equation for a weak base B is:

B + H2O ⇌ BH+ + OH-

The equilibrium constant Kb is used to determine the concentration of OH-. Similar to weak acids, an approximation method can be used when the base concentration is much larger than Kb.

Formula: $OH-$ ≈ √(Kb × initial base concentration)

An example is provided for calculating the pH of 250 mL of 0.250 M ammonia (NH3) with Kb = 1.75 × 10^-5:

Example: Using the quadratic equation or approximation method, $OH-$ is calculated, then pOH = -log$OH-$, and finally pH = 14 - pOH

This example illustrates the process of calculating pH for a weak base, considering the equilibrium and using the relationship between pH and pOH.

pH of Strong Acids

This section explains how to calculate the pH of strong acids. Strong acids fully dissociate in water, releasing hydrogen ions $H+$ or hydronium ions $H3O+$. The pH calculation depends on the number of protons released by the acid.

For monoprotic acids like HCl, which release one proton, the concentration of H3O+ equals the concentration of the acid. The pH is calculated using the formula:

Formula: pH = -log$H3O+$ = -log[acid concentration]

For diprotic acids like H2S, which release two protons, the concentration of H3O+ is twice the acid concentration. The pH formula becomes:

Formula: pH = -log[2 × acid concentration]

For polyprotic acids releasing n protons, the general formula is:

Formula: pH = -log[n × acid concentration]

An example is provided for calculating the pH of 100 mL of 0.1 M HCl:

Example: For 0.1 M HCl, pH = -log[0.1] = 1

This demonstrates how to apply the pH formula for a strong monoprotic acid.