$ 20 \; mL \; HCl \; 0.10 \; M $ are titrated by $ NaOH \; 0.050 \ M $. Calculate the pH after adding $ 50 \; mL \; NaOH $. First calculate the volume added at the E.P.:
Pour les réponses, utilisez (éventuellement plusieurs fois) les flèches ↑ et ↓ en haut! Terminez s.v.pl. cette question avant de passer à la suivante!
$V_e$ $=$ $\frac{0.020\cdot 0.10}{0.05}$ $=$ $40\;mL$
In which region of the titration are we and how will be the calculation?
After the E.P..: There is an excess of strong base $ NaOH $ added. The product $NaCl $ does not influence the pH.
Calculate the number of moles $ n_b $ base in excess!
One mole of acid reacted with one mole of base: $n_b$ $=$ $n_{added base\; base}$ $- $ $n_{initial\;acid}$ $n_b$ $=$ $0.050\cdot0.050$ $- $ $0.020\cdot0.10$ $=$ $5\cdot 10^{-4}\;mol$
Calculate the molarity $c_b$ of the base!
$c_b$ $=$ $\frac{n_b}{V_i+v}$ $c_b$ $=$ $\frac{5\cdot 10^{-4}}{0.070}=7.14\cdot 10^{-3}\;M$
ate the pH!
Strong base: $pH$ $=$ $14+log\;c_{b}$ $pH$ $=$ $14+log\;7.14\cdot 10^{-3}$ $=$ $11.9$
