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**Problem**: Determine Kb for the nitrite ion, NO 2−. In a 0.10-M solution this base is 0.0015% ionized.

###### FREE Expert Solution

###### FREE Expert Solution

Step 1:

Step 2:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{HNO}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\left[{{\mathrm{NO}}_{2}}^{-}\right]}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{}\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}}$

Step 3:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{ionization}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\left[{\mathrm{OH}}^{-}\right]}{{\mathbf{\left[}{{\mathbf{NO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}}_{\mathbf{initial}}}{\mathbf{\times}}{\mathbf{100}}}\phantom{\rule{0ex}{0ex}}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\mathbf{}\mathbf{=}{\mathbf{\left[}{{\mathbf{NO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}}_{\mathbf{initial}}\mathbf{}\mathbf{\times}\frac{\mathbf{\%}\mathbf{}\mathbf{ionization}}{\mathbf{100}\mathbf{}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{M}\mathbf{}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{0015}}{\mathbf{100}}$

**[OH ^{-}] = 1.5x10^{-6} M**

###### Problem Details

Determine K_{b} for the nitrite ion, NO _{2}^{−}. In a 0.10-M solution this base is 0.0015% ionized.

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