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At $25^{\circ} \mathrm{C}$, the dissociation constant of a base, $\mathrm{BOH}$, is $1.0 \times 10^{-12}$. The concentration of hydroxyl ions in $0.01 \mathrm{M}$ aqueous solution of the base would be:
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Verified Answer
The correct answer is:
$1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$
$$
\begin{gathered}
\mathrm{C}=0.01 \mathrm{M} \\
\mathrm{K}_b=1 \times 10^{-12} \text { at } 25^{\circ} \mathrm{C} \\
\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\
\mathrm{C} \quad 0 \quad 0
\end{gathered}
$$
at eq. $\mathrm{C}-\mathrm{C} \alpha \mathrm{C} \alpha \mathrm{C} \alpha$
$$
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } & =\mathrm{C} \alpha \\
{\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_b \mathrm{C}}=\sqrt{1 \times 10^{-12} \times 10^{-2}}
\end{aligned}
$$
$$
\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}
$$
Related Theory
The activities of the solutes are approximated with molarities. The dissociation constant is an immediate consequence of the law of mass action which describes equilibria in a more general way. The dissociation constant is also sometimes called ionization constant when applied to salts.
\begin{gathered}
\mathrm{C}=0.01 \mathrm{M} \\
\mathrm{K}_b=1 \times 10^{-12} \text { at } 25^{\circ} \mathrm{C} \\
\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\
\mathrm{C} \quad 0 \quad 0
\end{gathered}
$$
at eq. $\mathrm{C}-\mathrm{C} \alpha \mathrm{C} \alpha \mathrm{C} \alpha$
$$
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } & =\mathrm{C} \alpha \\
{\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_b \mathrm{C}}=\sqrt{1 \times 10^{-12} \times 10^{-2}}
\end{aligned}
$$
$$
\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}
$$
Related Theory
The activities of the solutes are approximated with molarities. The dissociation constant is an immediate consequence of the law of mass action which describes equilibria in a more general way. The dissociation constant is also sometimes called ionization constant when applied to salts.
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