A weak base has K_b = 1.5 × 10⁻⁹. What is the value of K_a for the conjugate acid?

Interpretation:

The Ka value for conjugate acid of a weak base has to be calculated by using the Kb of weak base.

Concept introduction:

A base (B) undergoes dissociation in an aqueous medium. The base dissocition gives ions BH+(aq.)  and OH−(aq.). The conjugate acid BH+(aq.)  may interact with H2O and forms BOH(aq.) and it can be represented as following equilibrium.

  B(aq.)+ H2O(l)⇌BH+(aq.) + OH−1(aq.)                                                 (1)

The dissociation constant for the acid is Kb,

  Kb=[BH+][OH−][B]

Interaction of conjugate acid BH+(aq.) with water is represented as,

  BH+(aq.)+ H2O(l)⇌B(aq.)+ H3O+(aq.)                                                 (2)

The dissociation constant for the conjugate acid is Ka,

  Ka=[B][H3O+][BH+]

From equation (1) and (2) the net equilibrium will be

  2H2O(l)⇌H3O+(aq.)+ OH−1(aq.)                                                             (3)

The equation (3) is known as the auto-ionization of water molecule. The dissociation constant for water is written as follows,

  Kw=[H3O+][OH−]

Thus from equation (1) and (2), there is an important relation established between ionization constant Kb for base and ionization constant Ka of conjugate acid i.e.

  Kw=Ka×Kb                                                                                               (4)

The ionisation constant of water has a constant value i.e. Kw=1×10-14. On taking negative logarithm of equation (4), a new relation established in terms of pKa,pKb and pKw as follows,

  -log(Kw)=-log(Ka×Kb)

   pKw= pKa+pKb                                                                                           (5)

The Ka value for conjugate acid (BH+ ) of the given weak base (B) is calculated below.

Given:

Value of Kb for weak base is 1.5×10−9.

Value of Kw for water is 1.0×10−14.

Using equation (4),Ka for conjugate acid is calculated as follows,

  Kw=Ka×Kb

Substitute the value of Kb and Kw

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