$ 20 \; mL \; HF \; 0.10 \; M $ are titrated by $ KOH \; 0.050 \; M $. Calculate the pH after adding $ 50 \; mL \; KOH $. Calculate first the volume added at the E.P.:
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$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 ?
After the E.P.: There is a strong base $ KOH $ excess added . The weak base $ F^- $ formed up to the point is neglected.
Calculate the number of moles $ n_b $ of strong base in excess !
One mole of acid reacted with one mole of base: $n_b$ $=$ $n_{added\; base}$ $- $ $n_{initial\;acid}$ $n_b$ $=$ $0.050\cdot0.050$ $- $ $0.020\cdot0.10$ $=$ $5\cdot 10^{-4}\;mol$
Calculate the molarity $c_b$ of the base!
$c_b$ $=$ $\frac{n_b}{V_i+v}$ $c_b$ $=$ $\frac{5\cdot 10^{-4}}{0.070}=7.14\cdot 10^{-3}\;M$
Calculate the pH !
Strong base: $pH$ $=$ $14+log\;c_{b}$ $pH$ $=$ $14+log\;7.14\cdot 10^{-3}$ $=$ $11.9$
pH during titration of a weak acid by a strong base
