pKa + pKb = pKw, Relationship between them?

The equation pK_a + pK_b = pK_w is based on the principle of acid-base chemistry and the ionization of water.

 It is a multipurpose equation that can be used to find the acid dissociation constant (K_a) or the base dissociation constant (K_b) using a fixed value of K_w at 25°C.

What are the components of pK_a + pK_b = pK_w? 

In pK_a + pK_b = pK_w :

• pK_a = negative logarithm to the base 10 of K_a.
• pK_a = -log₁₀ K_a where K_a = acid dissociation constant.

K_a measures the extent of ionization of an acid in an aqueous solution.

Acids are proton donors. A weak acid (HA) partially ionizes into H⁺ and A^– ions in water.

The greater the K_a value, the higher the acidity of the aqueous solution.

• pK_b = negative logarithm to the base 10 of K_b.
• pK_b = -log₁₀ K_b where K_b = base dissociation constant.

K_b measures the extent of ionization of a base in an aqueous solution.

Bases are proton acceptors. A weak base (B) partially ionizes in water to produce OH^– and BH⁺ ions.

The greater the K_b value, the higher the basicity of an aqueous solution.

• pK_w = negative logarithm to the base 10 of K_w.
• pK_w = -log₁₀ K_w where K_w = water dissociation constant.

At room temperature and pressure, the value of K_w stays fixed.

A water molecule dissociates completely to produce H⁺ and OH^– ions per H₂O molecule.

Therefore, 1 mole of H₂O dissociates to produce an equal concentration of hydrogen ions [H⁺] and hydroxide [OH^–] ions, i.e., [H⁺] = [OH^–] = 1 x 10^-7 mol/L.

K_w = [H⁺][OH^–]

∴ K_w = [H⁺][OH^–] = (1 x 10^-7)² = 1 x 10^-14 mol²/L².

More commonly used units for K_w are mol²/dm⁶ or mol²dm^-6, where 1 dm³ = 1 L.

The value of pK_w at 25°C is thus;

∴ pK_w = -log₁₀ K_w = -log₁₀ (1 x 10^-14) = 14

Hence;

∴pK_a + pK_b = 14 at 25°C.

Where and how to use pK_a + pK_b = pK_w? – Examples  

For a conjugate acid-base pair, if the value of either pK_a or pK_b is given, then we can easily apply the above formula to find the other unknown value. Both pK_a and pK_b are unitless entities, just like pK_w.

The smaller the pK_a value of a solution, the higher its pK_b value and vice versa, on a scale of 0-14.

Further insight: Very strong acids such as mineral acids (HCl, H₂SO₄, HNO₃) may have negative pK_a values or pK_a = 0 as these completely ionize in water, K_a = 1 gives pK_a = -log₁₀ (1) = 0. Similarly, strong bases (NaOH, KOH) have extremely low pK_b values.

Another point to remember is that the value pK_w = 14 at 25°C only, i.e., normal room temperature.

However, if the temperature changes, a new value of pK_w needs to be determined based on the given value of the water dissociation constant (K_w).

Let us see the following examples.

For example, The pK_a value of a substance X is 9.21; find its pK_b value at a fixed temperature of 25°C.

Solution

As per the question statement;

pK_a = 9.21

pK_b = ?

We already know that at 25°C;

pK_w = 14

So substituting all the above data into the equation pK_a + pK_b = pK_w and making pK_b the subject of the formula gives us:

⇒ pK_a + pK_b = pK_w

9.21 + pK_b = 14

pK_b = 14-9.21

∴ pK_b = 4.79

Result: The pK_b value of the substance X is 4.79.

Another example is– The K_b value for ammonia (NH₃), a weak base is 1 x 10²¹ mol/L. Use the formula (pK_a + pK_b = pK_w) to find the K_a value of ammonium (NH₄⁺) ions, the conjugate acid, at 25°C.

Solution

As the base dissociation constant (K_b) for ammonia is given in the question statement, so we can use it to find its pK_b as follows:

∴ K_b = 1 x 10²¹ mol/L

⇒ pK_b = -log₁₀ K_b

pK_b = -log₁₀ (1 x 10²¹)

pK_b = -21

At 25°C, K_w = 14. Substituting the above data into pK_a + pK_b = pK_w and making pK_w the subject of the formula gives us:

⇒ pK_a + pK_b = pK_w

pK_a + (-21) = 14

pK_a = 14 – (-21)  

pK_a = 35

Once we know the pK_a value, we can easily find the acid dissociation constant (K_a) as follows:

pK_a = -log₁₀ (K_a)

Taking the antilog of pK_a gives us K_a:

K_a = 10^-pKa

K_a = 10^-35

∴ K_a = 1 x 10^-35 mol/L

Result: The K_a for NH₄⁺ is 1 x 10^-35 mol/L at 25°C. 

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