$ 20 \; mL \; CH_2ClCOOH \; 0.10 \; M $ are titrated by $ NaOH \; 0.05 \; M $. Calculate the pH after adding $ 30 \; mL \; NaOH $. Calculate first the base volume added at the E.P. !
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$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 of the unreacted weak acid $ CH_2ClCOOH$ . The product $Na^++CH_2ClCOO^-$ contains the weak base $CH_2ClCOO^-$. We are in the presence of a buffer!
Calculate the number of moles $ n_a $ of acid that remains as well as the number of moles of base formed!
One mole of base has reacted with one mole of weak acid to form one mole of the corresponding weak base: $n_a$ $=$ $n_{initial}$ $- $ $n_{disappeared}$ $=$ $n_{initial}$ $- $ $n_{added \;base}$ $n_a$ $=$ $0.020 \cdot 0.10$ $- $ $0.030 \cdot 0.050$ $=$ $5\cdot 10^{-4}\;mol$ $n_b$ $= $ $n_{base \,ajoutée}$ $=$ $0,030\cdot 0,05$ $=$ $1,5\cdot 10^{-3}\;mol$
Calculate the pH !
Buffer: $pH$ $=$ $pK_a+log\frac{n_b}{n_a}$ $pH$ $=$ $3.75+log\frac{1.5\cdot 10^{-3}}{5\cdot 10^{-4}}$ $=$ $4.23$
pH during titration of a weak acid by a strong base
