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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