Dissociation Constants for Bases (HSC SSCE Chemistry): Revision Notes

Dissociation Constants for Bases

What are base dissociation constants?

When a weak base dissolves in water, only a fraction of its molecules split apart to form ions. This creates an equilibrium between the intact base molecules and the ions they produce. We use a special value called the base dissociation constant (represented as $K_b$) to measure how much splitting occurs.

The general equation showing a base reacting with water is:

$\text{B}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{BH}^+(\text{aq}) + \text{OH}^-(\text{aq})$

where B represents any base molecule.

The equilibrium expression for $K_b$ is:

$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$

Example: ammonia in water

Ammonia ($\text{NH}_3$) is a common weak base. When it reacts with water:

$\text{NH}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{NH}_4^+(\text{aq}) + \text{OH}^-(\text{aq})$

The $K_b$ expression becomes:

$K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = 1.8 \times 10^{-5}$

This tells us that at equilibrium, ammonia produces a relatively small concentration of ions compared to the amount that remains as $\text{NH}_3$ molecules.

How Kb relates to base strength

The size of $K_b$ directly indicates how strong a base is. Here's what the value tells you:

• Larger $K_b$ = Stronger base = More dissociation occurs
• Smaller $K_b$ = Weaker base = Less dissociation occurs

This works because a larger $K_b$ means the numerator (product concentrations) is bigger compared to the denominator (reactant concentration). In practical terms, more hydroxide ions ($\text{OH}^-$) are produced.

Strong bases vs weak bases

Strong bases, such as sodium hydroxide ($\text{NaOH}$), dissociate almost completely in water:

$\text{NaOH}(\text{aq}) \rightarrow \text{Na}^+(\text{aq}) + \text{OH}^-(\text{aq})$

Nearly 100% of the base splits apart, so the $K_b$ for strong bases is extremely large. For weak bases like ammonia, only a small percentage dissociates, giving much smaller $K_b$ values.

Common weak bases and their Kb values

The table below shows $K_b$ values for several weak bases at 25°C. Notice how the values range over several orders of magnitude, indicating very different base strengths:

Determining base concentration from pOH

The method you use to find base concentration depends on whether you're working with a strong or weak base.

Strong bases

For strong bases, dissociation is effectively 100% complete. This means:

• The final concentration of the intact base is negligible (essentially zero)
• The pOH directly tells you the original base concentration
• You can calculate $[\text{OH}^-]$ directly from pOH using: $[\text{OH}^-] = 10^{-\text{pOH}}$

Weak bases

For weak bases, the situation is more complex:

• Significant amounts of the base remain un-dissociated at equilibrium
• The pOH does NOT directly give you the base concentration
• You must use $K_b$ and the hydroxide ion concentration together to calculate the base concentration

The relationship between Ka and Kb

For any conjugate acid-base pair, there's a simple mathematical relationship connecting their equilibrium constants:

$K_a \times K_b = K_w$

where $K_w$ is the ion product of water ($1.0 \times 10^{-14}$ at 25°C).

Example: the NH₄⁺/NH₃ conjugate pair

Consider the ammonium ion ($\text{NH}_4^+$) and ammonia ($\text{NH}_3$), which form a conjugate acid-base pair.

For the weak acid $\text{NH}_4^+$:

$\text{NH}_4^+(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{NH}_3(\text{aq}) + \text{H}_3\text{O}^+(\text{aq})$

$K_a = 5.6 \times 10^{-10}$

For the corresponding weak base $\text{NH}_3$:

$\text{NH}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{NH}_4^+(\text{aq}) + \text{OH}^-(\text{aq})$

$K_b = 1.8 \times 10^{-5}$

Multiplying these values:

$K_a \times K_b = 5.6 \times 10^{-10} \times 1.8 \times 10^{-5} = 1.0 \times 10^{-14} = K_w$

Determining Kb experimentally

You can find the $K_b$ of a weak base by measuring the pH or pOH of a solution with known concentration, then using this data to calculate the hydroxide ion concentration.

The method

Starting with a base B that forms conjugate acid $\text{BH}^+$:

$\text{B}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{BH}^+(\text{aq}) + \text{OH}^-(\text{aq})$

The equilibrium expression is:

$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$

Because each base molecule that dissociates produces exactly one $\text{OH}^-$ ion and one $\text{BH}^+$ ion, we know that:

$[\text{OH}^-] = [\text{BH}^+]$

This allows us to simplify the expression to:

$K_b = \frac{[\text{OH}^-]^2}{[\text{B}]}$

Important assumptions

Required formulas

To convert between pH, pOH, and hydroxide concentration:

$[\text{OH}^-] = 10^{-\text{pOH}}$

$\text{pH} + \text{pOH} = 14$

Worked examples

Practical investigation: determining Kb for ammonia

Investigation 6.5 provides a hands-on method to experimentally determine the $K_b$ value for ammonia solution. The investigation uses dilution and pH measurements to gather data.

The dilution approach

Different concentrations of ammonia are prepared using serial dilution:

By measuring the pH of each solution and plotting pH against initial ammonia concentration, you can determine equilibrium concentrations of all species. This data then allows you to calculate $K_b$ using the equilibrium expression.

Remember!