MHT CET · Chemistry · Ionic Equilibrium
What is the pOH of millimolar solution of \(\mathrm{Ca}(\mathrm{OH})_2\) ?
- A 2.7
- B 10.3
- C 12.3
- D 11.3
Answer & Solution
Correct Answer
(A) 2.7
Step-by-step Solution
Detailed explanation
\(1 \mathrm{mM}=10^{-3} \mathrm{M} \)
\( \mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \)
\( 10^{-3} \mathrm{M} \)
\( \therefore 2 \times 10^{-3} \mathrm{M} \)
\( \therefore \mathrm{pOH}=-\) \(\log _{10}\left[\mathrm{OH}^{-}\right]=-\log _{10}\left(2 \times 10^{-3}\right)=2.7\)
\( \mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \)
\( 10^{-3} \mathrm{M} \)
\( \therefore 2 \times 10^{-3} \mathrm{M} \)
\( \therefore \mathrm{pOH}=-\) \(\log _{10}\left[\mathrm{OH}^{-}\right]=-\log _{10}\left(2 \times 10^{-3}\right)=2.7\)
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