$ 20 \; mL \; NaOH \; 0.10 \; M $ are titrated by $ HCl \; 0.050 \; M $. Calculate the pH after adding $ 50 \; mL \; HCl $. First calculate the volume added to E.P.:
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$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 ?
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_a $ of excess acid!
One mole of acid reacted with one mole of base: $n_a$ $=$ $n_{added\;acid}$ $- $ $n_{initial\;base}$ $n_a$ $=$ $0.050\cdot 0.050$ $- $ $0.020\cdot 0.10$ $=$ $5\cdot 10^{-4}\;mol$
Calculate the molarity $ c_a $ of the acid !
$c_a$ $=$ $\frac{n_a}{V_i+v}$ $c_a$ $=$ $\frac{5\cdot 10^{-4}}{0.070}=7.14\cdot 10^{-3}\;M$
Calculate the pH !
Strong acid: $pH$ $=$ $-log\;c_{acid}$ $pH$ $=$ $-log\;7.14\cdot 10^{-3}$ $=$ $2.15$
