Acid-base titration
Tutorial 8
pH during titration of a strong acid by a strong base
Schematic:
Determination of the base volume $ V_e $ added at the equivalent point
$n_{added\;base}$ $= $ $n_{initial\;acid}$
$V_e\cdot c_{base}$ $=$ $V_i\cdot c_{acid}$
so $V_e$ is calculated.
Calculation of pH ($v$ = volume of added base)
$v=0$
pH of a strong acid with molarity $c_{acid}$:
$pH=-log\;c_{acid}$
$v\lt V_e$
Determine the number of moles of acid $ n_a $ that have not yet reacted, then:
$pH$ $=$ $-log\frac{n_a}{V_i+v}$
$v=V_e$
$pH=7$
$v\gt V_e$
Determine the numbers of base moles $ n_b $ in excess, then:
$pH$ $=$ $14+log\frac{n_b}{V_i+v} $
$ 20 \; mL \; HCl $ are titrated by $ NaOH \; 0.1 \; M $. At the equivalent point, $ 25 \; mL \; NaOH $ were added.
Calculate the initial molarity of $ HCl $
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Complete please this question before moving on to the next one!
$V_e\cdot c_{base}=V_i\cdot c_{acide}$
$0.025\cdot 0.1$ $=$ $0.020\cdot c_{HCl}$
$c_{HCl}$ $=$ $0.125\;\frac{mol}{L}$
