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0.40 M H2NNH2 → weak base/conjugate base
0.80 M H2NNH3NO3
H2NNH3NO3 → H2NNH3+ + NO3-
H2NNH3+ → conjugate acid/weak acid
Kb of H2NNH2 = 3.0x10-6
Ka = 3.33x10-9
pKa = 8.48
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?
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.