Dissociation of acids and bases
Tutorial 3
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
$pK_a=-logK_a$
$K_a=10^{-pK_a}$
$pK_b=-logK_b$
$K_b=10^{-pK_b}$
$K_a\cdot K_b$ $=$ $10^{-14}$
$pK_a+pK_b$ $=$ $14$
Calculate the number of moles of $ HF $ and $ F^- $ present in $ 2.0 \; L $ of hydrofluoric acid $ 0.10 \; M $ at $pH=2.5$
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up!
Complete please this question before moving on to the next one!
First calculate the total number of moles (= initial number of moles of acid).
$n_{tot}$ $=$ $2.0\cdot 0.10$ $=$ $0.20\; mol$
Call $ x $ the number of moles of fluoride and enter in the expression of the ratio of the numbers of moles as a function of $ K_a $ and [H_3O^+] !
$\frac{[F^-]}{[HF]}=\frac{K_a}{[H_3O^+]}$
$\frac{n_{F^-}}{n_{HF}}=\frac{K_a}{[H_3O^+]}$
$\frac{x}{0.20-x}=\frac{10^{-3.17}}{10^{-2.5}}$
$x=0.214(0.20-x)$
$n_{F^-}$ $=$ $x$ $=$ $0.035\;mol$
$n_{HF}$ $=$ $0.20-x$ $=$ $0.165\;mol$
