Hòa tan hoàn toàn a gam CaO vào nước thu được 500 mL dung — Không quảng cáo

$CaO + {H_2}O \to Ca{(OH)_2}$

$\begin{array}{l}Ca{(OH)_2}\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,2HCl \to CaC{l_2}\,\,\, + 2{H_2}O\\\frac{{12,{{1.10}^{ - 3}}.0,1}}{2} \leftarrow 12,{1.10^{ - 3}}.0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol)\,\,\,\,\end{array}$

(a) $\Rightarrow {C_{M(Ca{{(OH)}_2})}}\, = \,\frac{{12,{{1.10}^{ - 3}}.0,1}}{{{{2.5.10}^{ - 3}}}}\, = 0,121(M)\,\,$

(b)

$\left. \begin{array}{l}CaO\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,{H_2}O \to Ca{(OH)_2}\\\frac{{12,{{1.10}^{ - 3}}.0,1}}{2} \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{12,{{1.10}^{ - 3}}.0,1}}{2}\,\,mol\end{array} \right\} \Rightarrow {m_{CaO}} = \frac{{12,{{1.10}^{ - 3}}.0,1}}{2}.56 = 3,388\,gam$

(c)

$\left. \begin{array}{l}Ca{(OH)_2} \to C{a^{2 + }} + 2O{H^ - }\\0,121\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,0,242\,M\end{array} \right\} \Rightarrow \left[ {{H^ + }} \right] = \frac{{{{10}^{ - 14}}}}{{0,242}} \Rightarrow pH =  - \lg \left[ {\frac{{{{10}^{ - 14}}}}{{0,242}}} \right] = 13,38$