Question

Daily Problem #26 Calculate the pH of a 2.00 M hydrazine (H2NNH.) solution. (K, for H2NNH, is 3.00 x 10).

Answer 1

Answer:

On dissolution in water, the following equilibrium will represent the basic reactions;

First we need to construct an ICE table for the calculation of equilibrium concentrations;

	H2NNH2 --->	N2H5+ +	OH-
Initial	2.00	0	0
Change	-a	+a	+a
Equilibrium	(2.00 - a)	a	a

The expression for Kb is as follows;

Kb = [N₂H₅⁺][OH^-]/[H₂NNH₂] = 3.00 x 10^-6

Kb = a²/(2.00 - a) = 3.00 x 10^-6

Or a² + 3.00 x 10^-6 a - 6.00 x 10^-6 = 0

On solving the above quadratic equation for a;

we will get a = 0.0024480 ~~ [OH^-]

and pOH = -log(0.0024480) = 2.61

and pH = 14 - pOH = 11.39

please let me know, if you have any doubts by commenting below the answer.

Thanks