Em tham khảo nha:
\(\begin{array}{l}
a)\\
{\rm{[}}O{H^ - }{\rm{]}} = 2{C_M}Ca{(OH)_2} = 2 \times 0,01 = 0,02M\\
b)\\
{n_{HCl}} = 0,1 \times 0,01 = 0,001\,mol\\
{n_{Ca{{(OH)}_2}}} = 0,1 \times 0,01 = 0,001\,mol\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{Ca{{(OH)}_2}}} > \dfrac{{{n_{HCl}}}}{2} \Rightarrow Ca{(OH)_2} \text{ dư } \\
{n_{Ca{{(OH)}_2}}} \text{ dư }= 0,001 - \dfrac{{0,001}}{2} = 0,0005\,mol\\
{n_{O{H^ - }}} = 2{n_{Ca{{(OH)}_2}}} \text{ dư }= 0,001\,mol\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,001}}{{0,1 + 0,1}} = 0,005M
\end{array}\)
