$ 20 \; mL \; HCl \; 0.10 \; M $ are titrated by $ NaOH \; 0.05 \; M $. Calculate the pH after adding $ 30 \; mL \; NaOH $. First calculate the base volume added at E.P.:
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$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 strong acid $HCl $ which has not reacted. The product $NaCl $ does not influence the pH.
Calculate the number of moles $ n_a $ of acid that remain !
One mole of base was reacted with one mole of acid: $n_a$ $=$ $n_{initial}$ $- $ $n_{disappeared}$ $=$ $n_{initial}$ $- $ $n_{added \; base}$ $n_a$ $=$ $0.020\cdot 0.10$ $- $ $0.030\cdot 0.050$ $=$ $5\cdot 10^{-4}\;mol$
Calculate the $ c_a $ molarity of the remaining acid !
$c_a$ $=$ $\frac{n_a}{V_i+v}$ $c_a$ $=$ $\frac{5\cdot 10^{-4}}{0.050}=0.01\;M$
Calculate the pH !
Strong acid: $pH$ $=$ $-log\;c_a$ $pH$ $= $ $-log\;0.010$ $=$ $2.0$
