0.40 M H_{2}NNH_{2} → weak base/conjugate base

0.80 M H_{2}NNH_{3}NO_{3}

H_{2}NNH_{3}NO_{3} → H_{2}NNH_{3}^{+} + NO_{3}^{-}

H_{2}NNH_{3}^{+} → conjugate acid/weak acid

**Calculate pK _{a}:**

K_{b} of H_{2}NNH_{2} = 3.0x10^{-6}

$\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{a}}\mathbf{\xb7}\overline{){\mathbf{K}}_{\mathbf{b}}}}{\overline{){\mathbf{K}}_{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}}$

**K _{a} = 3.33x10**

$\overline{){\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{a}}}\phantom{\rule{0ex}{0ex}}{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{)}$

**pK _{a} = 8.48**

**Calculate pH:**

Calculate the pH of a solution that is 0.40 M H_{2}NNH_{2} and 0.80 M H_{2}NNH_{3}NO_{3}. In order for this buffer to have pH = pK_{a}, would you add HCl or NaOH? What quantity (moles) of which reagent would you add to 1.0 L of the original buffer so that the resulting solution has pH = pK_{a}?

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