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### Topic: How do I calculate the pH of a 0.1 M solution of (N2H5) (HSO4)?  (Read 1221 times)

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#### dolphinsea14

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##### How do I calculate the pH of a 0.1 M solution of (N2H5) (HSO4)?
« on: June 22, 2024, 06:34:42 PM »
Hydrazine (N2H4) is a weak diprotic base (Kb1 = 8.5∙10–7; Kb2 = 8.9∙10–16). It can form

various anhydrous salts with sulfuric acid. The second acid dissociation constant of

sulfuric acid is Ka2 = 1.02∙10–2.
pH should be 1.57. I obtained this correct value, but I didn’t get the same cubic equation. In my equation the first member in the second bracket 𝐾𝑏2/𝐾𝑤 is missing. I took into consideration the hydrolysis of N2H5+, dissociation of HSO4- and autoprotolysis of water, but I am missing something.

#### Borek

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##### Re: How do I calculate the pH of a 0.1 M solution of (N2H5) (HSO4)?
« Reply #1 on: June 23, 2024, 02:34:39 AM »
Hard to comment not seeing the derivation.

Technically there 5 equilibria present - two for the acid, two for the base, one for water, so there are 5 equations that needs to be combined, and the final equation should be not cubic, but of 5th degree. In this case of these equilibria one for the acid and one for the base can be safely neglected if all you are interested in is a practical result, not a theoretical formula.
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#### dolphinsea14

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##### Re: How do I calculate the pH of a 0.1 M solution of (N2H5) (HSO4)?
« Reply #2 on: July 15, 2024, 03:15:57 PM »
This is an IChO problem. You can see it in the attached pdf. There is no derivation, just the final cubic equations are given. Please help me with the derivation, if you understand how it is obtained.

#### Borek

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##### Re: How do I calculate the pH of a 0.1 M solution of (N2H5) (HSO4)?
« Reply #3 on: July 15, 2024, 04:27:50 PM »
From the description they didn't use all possible equations, assuming concentration of H2SO4 and N2H4 to be negligible (thus no equilibrium for N2H4/N2H5+ and H2SO4/HSO4-). Try to write all equations mentioned and solve, that should yield 3rd degree polynomial in [H+].

I haven't tried to derive (mainly because I don't like combining Ka and Kb, I prefer to use just Ka for everything, IMHO it makes math easier), but answers given (pH 1.56 and 4.83) definitely look correct.
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