$ 0.300 \; g \; CH 3COOH$ are dissolved in water forming $50\; ml$ of solution. $ 10 \; mL $ of this solution are titrated by $ NaOH \; 0.050 \ M $. Calculate the pH after adding $ 20 \; mL \; NaOH $. Calculate first the initial molarity of $ CH_3COOH$
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$c_{CH_3COOH}$ $=$ $\frac{0.300}{60\cdot 0.050}$ $=$ $0.10\;M$
Calculate the volume added at the E.P. !.
$V_e$ $=$ $\frac{0.010\cdot 0.10}{0.050}$ $=$ $20\;mL$
So we are at the E.P.. What do we have and how do we calculate the pH ?
We have only $CH_3COO^-+Na^+$. It will be necessary to calculate the pH of the weak base $CH_3COO^-$.
Calculate the molarity $ c_b $ of this base!
$n_b$ $=$ $n_{initial \;acid}$ $=$ $0.10\cdot 0.05$ $=$ $0.005\; mol$. $c_b$ $=$ $\frac{n_b}{V_i+V_e}$ $=$ $\frac{0.005}{0.050+0.020}$ $=$ $0.071 M$
Calculate the pH of this weak base by a second degree equation !
Let $x$ $=$ $[OH^-]$. $x^2+c_bx-c_bK_b$ $=$ $0$ $x^2+0.071\cdot x-0.071\cdot 10^{9.25}$ $=$ $0$ $x$ $=$ $6.32\cdot 10^{-6}\;M$ $pOH$ $=$ $-log(6.32\cdot 10^{-6}) =5.20$ $pH$ $=$ $14-5.20$ $=$ $8.80$
pH during titration of a weak acid by a strong base
