Dissociation of acids and bases
Tutorial 2
Dissociation of acids and bases ; acidity and basicity constants
Dissociation of a strong acid $HB $
$HB + H_2O $ $\rightarrow$ $H_3O^+ + B^-$
Dissociation of a strong base $B $
$B + H_2O $ $\rightarrow $ $OH^- + HB $
(Charges may differ!)
Dissociation of a weak acid $HB$
$HB + H_2O$ $\leftrightarrows$ $H_3O^+ + B^-$
(Charges may differ!)
Constant of acidity
($\neq $ constant of the former equilibrium):
$K_a = \frac{[H_3O^+] [B^-]} {[HB]} $
Dissociation of a weak base $B $
$B + H_2O $ $\leftrightarrows $ $OH^- + HB^+ $
(Charges may differ!)
Basicity Constant
($\neq $constant of the former equilibrium):
$K_b=\frac{[OH^-][HB]}{[B]}$
Relations
$pK_a=-logK_a$
$K_a=10^{-pK_a}$
$pK_b=-logK_b$
$K_b=10^{-pK_b}$
$K_a\cdot K_b$ $=$ $10^{-14}$
$pK_a+pK_b$ $=$ $14$
For a given weak acid, find a simple relationship that lets you know under which conditions it has as many moles of acid as the corresponding base!
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up!
Complete please this question before moving on to the next one!
From the definition of $ K_a $.
$ K_a $ $ = $ $ \frac{[B^-] [H_3O^+]} {[HB]} $
Isolate in $ K_a $ the ratio of the moles of acid and base species!
$ \frac{\frac{n_{B^-}}{V}} {\frac{n_{HB}}{V} } $
What happens if $ n_{B^-} = n_{HB} $?
$\frac{K_a}{[H_3O^+]}=1$
$K_a$ $=$ $[H_3O^+]$
$pK_a$ $=$ $pH$
