# Problem: Calculate the pH of a solution that is 0.40 M H2NNH2 and 0.80 M H2NNH3NO3. In order for this buffer to have pH = pKa, 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 = pKa?

###### FREE Expert Solution

0.40 M H2NNH2 → weak base/conjugate base

0.80 M H2NNH3NO3

H2NNH3NO3 → H2NNH3+ + NO3-

H2NNH3+ → conjugate acid/weak acid

Calculate pKa:

Kb of H2NNH2 = 3.0x10-6

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

Ka = 3.33x10-9

pKa = 8.48

Calculate pH:

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###### Problem Details

Calculate the pH of a solution that is 0.40 M H2NNH2 and 0.80 M H2NNH3NO3. In order for this buffer to have pH = pKa, 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 = pKa?