$pH$ of mixtures
Tutorial 4
pH of a mixture of acids
Strong acid 1 (volume $ V_1 $) / strong acid 2 (volume $ V_2 $)
We calculate:
- the numbers of moles $ n_1 $ and $ n_2 $
so:
$ pH = -log\frac{n_1 + n_2}{V_1 + V_2} $
Strong acid 1 (volume $ V_1 $) / weak acid 2 (volume $ V_2 $)
We calculate:
The number of moles of strong acid $ n_1 $
Then, neglecting the weak acid, but not the dilution:
$ pH = -log \frac {n_1}{V_1 + V_2} $
Complete treatment
If both acids are weak, a simple treatment is not indicated!
What is the pH of a mixture of $ 0.50 \; L \; HCl \; 1.0 \; \frac{mol}{L} $ with $ 500 \; mL \; HF \; $ $ 5 \; \% $?
Calculate the number of moles $HCl$ !
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up!
Complete please this question before moving on to the next one!
$n_1=n_{HCl}$ $=$ $0.50\cdot 1.0$ $=$ $0.50 \;mol$
Calculate the pH (Reminder: $ HF $ weak acid !)
$pH=-log\frac{n_1}{V_1+V_2}=-log\frac{0.50}{0.50+0.50}=0.30$
