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Ionisation constant of Weak Acid and Bases




Ionisation constant of Weak Acid

  • Consider an Weak Acid HA and its dissociation as given below
    $HA + H_2O \rightleftharpoons A^- + H_3O^+$
  • The ionisation constant or the dissociation constant of Weak Acid HA is defined as
    $K_a=\frac {[H^+][A^-]}{[HA]}$
    or $K_a=\frac {[H_3O^+][A^-]}{[HA]}$
  • $K_a$ is a measure of the strength of the acid HX at a given temperature i.e., larger the value of Ka, the stronger is the acid.
  • $K_a$ is a dimensionless quantity with the understanding that the standard state concentration of all species is 1M

Formula of Ionisation constant of Weak Acid

$HA + H_2O \rightleftharpoons A^- + H_3O^+$
Initial Concentration At time=0
c       0        0
Let $\alpha$ be the extent of ionization
Change concentration
$-c\alpha$ $c\alpha$ $c\alpha$
Equilibrium Concentration
$c -c\alpha$        $c\alpha$        $c\alpha$
So, $K_a = \frac {[H^+][A^-]}{[HA]}$
$ = \frac{c\alpha \times c\alpha}{c(1 - \alpha)}=\frac{c \alpha^2}{1-\alpha }$

How to find the pH of weak Acid

  • Step 1. The species present before dissociation are identified as Brönsted-Lowry acids / bases.
  • Step 2. Balanced equations for all possible reactions i.e., with a species acting both as acid as well as base are written.
  • Step 3. The reaction with the higher Ka is identified as the primary reaction whilst the other is a subsidiary reaction.
  • Step 4. Enlist in a tabular form the following values for each of the species in the primary reaction
    (a) Initial concentration, c.
    (b) Change in concentration on proceeding to equilibrium in terms of $\alpha$, degree of ionization.
    (c) Equilibrium concentration.
  • Step 5. Substitute equilibrium concentrations into equilibrium constant equation for principal reaction and solve for $\alpha$.
  • Step 6. Calculate the concentration of species in principal reaction.
  • Step 7. Calculate $pH = – log[H_3O^+]$

Ionisation constant of Weak Base

  • Consider an Weak Acid HA and its dissociation as given below
    $MOH \rightleftharpoons M^+ + OH^- $
  • The ionisation constant or the dissociation constant of Weak Bases MOH is defined as
    $K_b=\frac {[M^+][OH^-]}{[MOH]}$
  • $K_b$ is a measure of the strength of the Bases HX at a given temperature i.e., larger the value of Kb, the stronger is the Bases.
  • $K_b$ is a dimensionless quantity with the understanding that the standard state concentration of all species is 1M
  • The pH scale for the hydrogen ion concentration has been extended to get:
    $pK_b = –log (K_b)$

Formula of Ionisation constant of Weak Bases

$MOH \rightleftharpoons M^+ + OH^- $
Initial Concentration At time=0
c       0        0
Let $\alpha$ be the extent of ionization
Change concentration
$-c\alpha$        $c\alpha$       $c\alpha$
Equilibrium Concentration
$c -c\alpha$        $c\alpha$       $c\alpha$
So, $K_a = \frac {[H^+][A^-]}{[HA]}$
$ = \frac{c\alpha \times c\alpha}{c(1 - \alpha)}=\frac{c \alpha^2}{1-\alpha }$

Relation between Ka, Kb & Kw

  • Ka and Kb represent the strength of an acid and a base, respectively.
  • They are related in a simple manner in conjugate acid-base pair
$HA + H_2O \rightleftharpoons A^- + H_3O^+$
$K_a = \frac{[A^-].[H_3O^+]}{[HA][H_2O]}$
$A- + H_2O \rightleftharpoons HA + OH^-$
$K_b = \frac {[HA][OH^-]}{[A^-][H2O]}$
Now,
$K_a. K_b = \frac{[A^-].[H_3O^+]}{[HA][H_2O]} \times \frac {[HA][OH^-]}{[A^-][H2O]}$
$K_a. K_b = [H_3O^+] [OH^-] = K_w$
Ka. Kb = Kw
if we take negative logarithm of both sides of the equation, then pK values of the conjugate acid and base are related to each other by the equation:
$pK_a + pK_b = pK_w = 14$


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