MHT CET · Chemistry · Ionic Equilibrium
The number of hydroxyl ions in \(10 \mathrm{~cm}^{3}\) of \(0.2 \mathrm{M}\) HCl solution is
- A \(5 \times 10^{-14}\)
- B \(3 \times 10^{9}\)
- C \(3 \times 10^{12}\)
- D \(5 \times 10^{-12}\)
Answer & Solution
Correct Answer
(C) \(3 \times 10^{12}\)
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{H}^{+}\right]=0.2 \mathrm{M}\) (given, \(\mathrm{HCl}=0.2 \mathrm{M}\) )
\(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}\)
\(\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{0.2}=5 \times 10^{-14} \mathrm{M}\)
Concentration of hydroxyl ions in \(0.2 \mathrm{M} \mathrm{HCl}\)
\(
=5 \times 10^{-14} \mathrm{M}
\)
\(\therefore\) The number of moles of hydroxyl ions
\(
=5 \times 10^{-14} \times 1000
\)
The number of \(\mathrm{OH}^{-}\) in \(10 \mathrm{~cm}^{3}\)
\(
\begin{array}{l}
=\frac{5 \times 10^{-14} \times 1000 \times 6.023 \times 10^{23}}{10} \\
=3 \times 10^{12}
\end{array}
\)
\(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}\)
\(\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{0.2}=5 \times 10^{-14} \mathrm{M}\)
Concentration of hydroxyl ions in \(0.2 \mathrm{M} \mathrm{HCl}\)
\(
=5 \times 10^{-14} \mathrm{M}
\)
\(\therefore\) The number of moles of hydroxyl ions
\(
=5 \times 10^{-14} \times 1000
\)
The number of \(\mathrm{OH}^{-}\) in \(10 \mathrm{~cm}^{3}\)
\(
\begin{array}{l}
=\frac{5 \times 10^{-14} \times 1000 \times 6.023 \times 10^{23}}{10} \\
=3 \times 10^{12}
\end{array}
\)
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