$ 20 \; mL \; NaOH \; 0.10 \; M $ are titrated by $ HCl \; 0.05 \ M $. Calculate the pH after adding $ 30 \; mL \; HCl $. First calculate the volume added to the E.P.:
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up! Complete please this question before moving on to the next one!
$V_e$ $=$ $\frac{0.020\cdot 0.10}{0.05}$ $=$ $40\;mL$
In which region of the titration are we and how will be the calculation?
Before the E.P.: There remains the strong $ NaOH $ base that has not reacted. The product NaCl does not influence the pH.
Calculate the number of moles $ n_b $ of base that remains!
One mole of acid reacted with one mole of base: $n_b$ $=$ $n_{initial}$ $- $ $n_{disappeared}$ $=$ $n_{initial}$ $- $ $n_{added\; acid}$ $n_b$ $=$ $0.020\cdot 0.10$ $- $ $0.030\cdot 0.050$ $=$ $5\cdot 10^{-4}\;mol$
Calculate the molarity $ c_b $ of the base that remains!
$c_b$ $=$ $\frac{n_b}{V_i+v}$ $c_b$ $=$ $\frac{5\cdot 10^{-4}}{0.050}=0.01\;M$
Calculate the pH !
Strong base: $pH$ $=$ $14+log\;c_{base}$ $pH$ $=$ $14$ $+$ $log\;0.010$ $=$ $12$
