Practice calculating ion concentrations in a dibasic aqueous solution. Hydrazine

Question

Practice calculating ion concentrations in a dibasic aqueous solution. Hydrazine, N_2H_4, is dibasic and has K_b1 = 8.51 times 10^-7 and K_b2 = 8.91 times 10^-16. Calculate the concentration of OH^- ions in a 0.36 M aqueous solution of hydrazine. Express your answer with three significant figures. This question will be shown after you complete previous question(s). This question will be shown after you complete previous question(s). What is the pH of a 0.36 M aqueous solution of hydrazine? Express your answer with three significant figures.

Explanation / Answer

we consider only Kb1 since Kb2 is very low and thus we have low OH- production by 2nd dissociation eq

N2H4 (aq) + H2O (l) <---> N2H5+ (aq) + OH- (aq)   

at equilibrium [N2H4] = 0.36-X , [N2H5+] =[OH-] = X ,

Kb1 = [N2H5+] [OH-] /[N2H4]

8.51 x 10^-7 = (X) (X) / ( 0.36-X)

X^2 + ( 8.51 x 10^-7) X - 3.06 x 10^-7 = 0

X = 5.53 x 10^-4 M

[OH-] = 5.53 x 10^- 4M

D) pOH = -log [OH-]

   = -log ( 5.53 x 10^-4)

        = 3.26

pH = 14 - pOH

       = 14 - 3.26

       = 10.7

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