$pH$ of mixtures
Tutorial 5
pH of a mixture of bases
Strong base 1 (volume $ V_1 $) / strong base 2 (volume $ V_2 $)
We calculate:
- the numbers of moles $ n_1 $ and $ n_2 $ of ions $ OH^- $ released by the bases
so:
$ pOH $ $ = $ $ -log \frac{n_1 + n_2}{V_1 + V_2} $
$ pH $ $ = $ $ 14-pOH $
Strong base 1 (volume $ V_1 $) / weak base 2 (volume $ V_2 $)
We calculate:
The number of moles of ions $ OH^- $ released by the strong base: $ n_1 $
Then, neglecting the weak base, but not the dilution:
$ pOH = -log\frac{n_1}{V_1 + V_2} $
$ pH $ $ = $ $ 14-pOH $
Complete treatment
If both bases are weak, a simple treatment is not indicated!
What is the pH of a mixture of $ 2.0 \; L \; NH_3 \; 1.0 \; \frac{mol}{L} $ with $ 8.0 \; L \; NaOH \; 0.5 \; \frac{mol}{L} $?
Calculate the number of moles of $NaOH$!
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up!
Complete please this question before moving on to the next one!
$n_{NaOH}$ $=$ $8.0\cdot 0.5$ $=$ $4.0\; mol$
$n_1=n_1(OH^{-})$ $=$ $8.0\cdot 0.5$ $=$ $4.0\; mol$
Calculate the pH. ($NH_3$ weak base !)
$pOH$ $=$ $-log\frac{n_1}{V_1+V_2}$ $=$ $-log\frac{4.0}{2.0+8.0}$ $=$ $0.40$
$pH$ $=$ $14-0.40$ $=$ $13.60$
