Acid-base titration
Tutorial 21
pH during titration of a weak base by a strong acid
Schematic:
Determining the acid volume $ V_e $ added at the equivalent point
$ n_{added \;acid } = n_{initial \; base} $
$ V_e \cdot c_{acid} = V_i \cdot c_{base} $
Hence $ V_e $
pH by volume $ v $ of base added
$ v = 0 $
pH of a weak base of molarity $ c_{base} $:
$ x = [OH^-] $
$ x ^ 2 + c_{base} x-c_{base} K_a = 0 $
etc...
$v\lt V_e$
Determine the numbers of weak base moles $ n_b $ that have not yet reacted as well as the number of moles of weak acid $ n_b $ formed.
pH of buffer !
$pH=pK_a+log\frac{n_b}{n_a}$
$v=V_e$
pH of a weak acid $c_a$:
The number of moles of weak acid present at this time $n_a$ $=$ $n_{added \; acid}$ $=$ $ n_{initial\;base}$
Then: $c_a=\frac{n_a}{V_i+V_e}$
and with
$x=[H_3O^+$:
$x^2+c_ax-c_aK_a=0$
etc..
$v\gt V_e$
Determine the numbers of moles of strong acid $ n_a $ in excess
$pH=-log\frac{n_a}{V_i+v} $
$ 20 \; mL \; CH_3NH_2 \; 0.10 \; M $ are titrated by $ HCl \; 0.1 \; M $. Calculate the pH at the beginning of the titration.
$ CH_3NH_2 $ is what kind of base ?
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up!
Complete please this question before moving on to the next one!
Calculate $ [OH^-] $ by a second degree equation!
Let $x$ $=$ $[OH^-]$
$x^2+c_bx-c_bK_b$ $=$ $0$
$x^2+0.10\cdot x-0.10\cdot 10^{3.30}$ $=$ $0$
$x=7.08\cdot 10^{-3}\;M$
Then calculate the pOH and the pH!
$pOH$ $=$ $-log[OH^-]$ $=$ $-log(7.08\cdot 10^{-3})$ $=$ $2.15$
$pH$ $=$ $14-pH$ $=$ $11.85$
