What is the pH of a mixture of $ 2.0 \; L \; Ca(OH)_2 \; 1.0 \; \frac{mol}{L} $ with $ 8.0 \; L \; NaOH 1\; 0.5 \; \frac{mol}{L} $? Calculate the numbers of moles!
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$n_{Ca(OH)_2}$ $=$ $2.0\cdot 1.0$ $=$ $2.0\; mol$ $n_1=n_1(OH^{-})$ $=$ $2\cdot 2.0\cdot 1.0$ $=$ $4.0\; mol$ $n_{NaOH}$ $=$ $8.0\cdot 0.5$ $=$ $4.0\; mol$ $n_2=n_2(OH^{-})$ $=$ $8.0\cdot 0.5$ $=$ $4.0\; mol$
Calculate the pH !
$pOH$ $=$ $-log\frac{n_1+n_2}{V_1+V_2}$ $=$ $-log\frac{4.0+4.0}{2.0+8.0}$ $=$ $-0.10$ $pH$ $=$ $14-(-0.10)$ $=$ $14.10$
