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multiply the ka and kb expressions from 1a and 1b together and simplify the result. why is this result significant and what is the numerical value?

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Final answer:

The product of the acid dissociation constant (Ka) and the base dissociation constant (Kb) equals the ionic product of water (Kw), and at 25 °C, Kw is 1.0 x 10⁻¹⁴. This relationship is crucial in acid-base chemistry and helps predict the strength of conjugate acid-base pairs.

Step-by-step explanation:

When multiplying the Ka (acid dissociation constant) and Kb (base dissociation constant) together, we find a relationship significant to acid-base chemistry. The product of Ka and Kb for a conjugate acid-base pair is a constant at a given temperature, known as the ionic product of water (Kw). This is expressed in the equation pKa + pKb = pKw. For water at 25 °C, Kw is always 1.0 x 10⁻¹⁴. This implies that if Ka is small, indicating a weak acid with little dissociation, then Kb must be large, describing a strong conjugate base.

To demonstrate this, consider a hypothetical scenario where Ka = 2.0 x 10⁻¹⁼⁵ and Kb = 6.4. Multiplying these values gives K3 (the equilibrium constant for the net reaction) as follows: K3 = Ka x Kb = (2.0 x 10⁻¹⁼⁵) x 6.4 = 1.3 x 10⁻¹⁵, conforming to the principle where the equilibrium constant for a reaction is the product of the equilibrium constants for the sum of that reaction's individual steps.

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