Acid-base titration
Tutorial 5
pH during titration of a strong base with a strong acid
Schematic:
Determination of the volume of acid $ V_e $ added at the equivalent point
At E.P.:
$n_{acide\;added}= n_{base\;initial}$
$V_e\cdot c_{acid}=V_i\cdot c_{base}$
so $V_e$ is found.
pH according to the volume $ v $ of added acid
$v=0$
pH of a strong base of molarity $c_{base}$:
$pH=14+logc_{base}$
$v\lt V_e$
Determine the numbers of base moles $ n_b $ that have not yet reacted, then:
$pH=14+log\frac{n_b}{V_i+v}$
$v\;=\;V_e$
$pH=7$
$v\gt V_e$
Determine the number of moles of acid $ n_a $ in excess
$pH=-log\frac{n_a}{V_i+v} $
$ 0.196 \; g \; KOH$ are dissolved in water forming $ 50\; ml$ of solution.
$ 10 \; mL $ of this solution are titrated by $ HCl \; 0.050 \; M $.
Calculate the pH after adding $ 20 \: mL \: HCl $.
Calculate first the initial molarity of $ KOH $!
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Complete please this question before moving on to the next one!
$c_{KOH}$ $=$ $\frac{0.196}{39.1\cdot 0.050}$ $=$ $0.10\;M$
Calculate the volume added to the E.P.:. !.
$V_e$ $=$ $\frac{0.010\cdot 0.10}{0.050}$ $=$ $20\;mL$
We are at the E.P., so:
$pH=7.0$
