# Calculating dissociation constant at different pKw values

What are the standard approaches to calculating how dissociation constants for acids and bases, $$\mathrm{p}K_\mathrm{a}$$ and $$\mathrm{p}K_\mathrm{b}$$, change as the ionic product of water, $$\mathrm{p}K_\mathrm{w}$$, changes?

• They primarily change with temperature, but they still follow pKa + pKb = pKw – Poutnik Jun 28 '20 at 20:06
• Is the change generally "linear", if pKw decreases, do pKa and pKb both decrease proportionally (pKa + pKb = pKw)? – Asker123 Jun 28 '20 at 20:20
• @Asker123: Show first what you think in details. – Mathew Mahindaratne Jun 28 '20 at 20:21
• I don't really know how you'd think anyone could show more what they think than "if the equation is pKa + pKb = pKw, will both acid and conjugate base disassociation constant generally change "linearly"" – Asker123 Jun 28 '20 at 20:47

The equation

$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}\tag{1}$$

has two degrees of freedom, so two values are independent on each other and the third one depends on the other two.

Reaction equilibrium constant $$K_\mathrm{a}$$ for

$$\ce{HA + H2O <=> H3O+ + A-}\tag{R1}$$

is chemically independent on $$K_\mathrm{w}$$ for reaction

$$\ce{2 H2O <=> H3O+ + OH-}\tag{R2}$$

OTOH, the equilibrium constant $$K_\mathrm{b}$$ for the conjugate base is derived from the above two.

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{K_\mathrm{w}[\ce{A-}]}{[\ce{HA}][\ce{OH-}]} = \frac{K_\mathrm{w}}{K_\mathrm{b}}\tag{2}$$

$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$$

Another thing is that both $$\mathrm{p}K_\mathrm{a}$$ and $$\mathrm{p}K_\mathrm{w}$$ depend on temperature. It is not dependence

$$\Delta T \to \Delta \mathrm{p}K_\mathrm{w} \to \Delta \mathrm{p}K_\mathrm{a},\tag{3}$$

but

$$\begin{cases}\tag{4} \Delta T \to \Delta \mathrm{p}K_\mathrm{w} \\ \Delta T \to \Delta \mathrm{p}K_\mathrm{a} \end{cases}$$

For the temperature dependence of equilibrium constants, there is van 't Hoff equation, which calculates the constant value changes from the reaction enthalpy:

$$\ln \frac{K_2}{K_1} = \frac{-\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right),$$

from which:

$$K_2 = K_1 \cdot \exp{\left( \frac{-\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\right)}\tag{5}$$