MCQ
The $pH$ at which $\mathrm{Mg}(\mathrm{OH})_2\left[\mathrm{~K}_{\mathrm{sp}}=1 \times 10^{-11}\right]$ begins to precipitate from a solution containing $0.10 \ \mathrm{M}$ $\mathrm{Mg}^{2+}$ ions is_____
- A$08$
- ✓$09$
- C$10$
- D$11$
${\left[\mathrm{Mg}^{2+}\right][\mathrm{OH}]^2=10^{-11}}$
$0.1 \times\left[\mathrm{OH}^{-}\right]^2=10^{-11} \Rightarrow\left[\mathrm{OH}^{-}\right]=10^{-5}$
$\Rightarrow \mathrm{pOH}=5 \quad \Rightarrow \mathrm{pH}=9$
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$(1)$ $\begin{array}{*{20}{c}}
{C{H_3}CH = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(2)$ $C{H_3}CH = CHC{H_3}$
$(3)$ $\mathop {C{H_3}C}\limits^{\begin{subarray}{l}
\,\,\,\,\,\,\,\,{\begin{array}{*{20}{c}}
{}&H
\end{array}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,|
\end{subarray} } = CHC{H_2}C{H_3}$
$(4)$ $\begin{array}{*{20}{c}}
{C{H_3}C = C - C{H_3}} \\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_{3\,\,}}\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$