Exercise 23.95 When hydrazine is dissolved in water it acts like a base. N2H4( a

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Exercise 23.95

When hydrazine is dissolved in water it acts like a base.
N2H4(aq)+H2O(l)N2H+5(aq)+OH(aq)
Kb1=8.5×107
N2H+5(aq)+H2O(l)N2H2+6(aq)+OH(aq)
Kb2=8.9×1016

Part A

Calculate the Kb for the overall reaction of hydrazine forming N2H2+6.

Express your answer using two significant figures.

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Part B

Calculate Ka1 for N2H+5.

Express your answer using two significant figures.

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Part C

Calculate the concentration of hydrazine in a solution buffered at a pH of 8.5 for a solution that was made by dissolving 1.2×102 mol of hydrazine in 1 L of water.

Express your answer using one significant figure.

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Part D

Calculate the concentration of N2H+5 in this solution.

Express your answer using two significant figures.

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Part E

Calculate the concentration of N2H2+6 in this solution.

Exercise 23.95

When hydrazine is dissolved in water it acts like a base.
N2H4(aq)+H2O(l)N2H+5(aq)+OH(aq)
Kb1=8.5×107
N2H+5(aq)+H2O(l)N2H2+6(aq)+OH(aq)
Kb2=8.9×1016

Part A

Calculate the Kb for the overall reaction of hydrazine forming N2H2+6.

Express your answer using two significant figures.

Kb =

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Part B

Calculate Ka1 for N2H+5.

Express your answer using two significant figures.

Ka1 =

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Part C

Calculate the concentration of hydrazine in a solution buffered at a pH of 8.5 for a solution that was made by dissolving 1.2×102 mol of hydrazine in 1 L of water.

Express your answer using one significant figure.

[N2H4] =   M  

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Part D

Calculate the concentration of N2H+5 in this solution.

Express your answer using two significant figures.

[N2H+5] =   M  

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Part E

Calculate the concentration of N2H2+6 in this solution.

Explanation / Answer

A)

Kb overall:

N2H4 + H2O <--> N2H5+ + OH-

N2H5+ + H2O <--> N2H6+2 + OH-

Koveral = [N2H6+2][OH-]^2 /[N2H4]

Koveral = Kb1*Kb2 = (8.5×107)(8.9×1016) = 7.565*10^-22

B)

Calculate Ka1 for N2H5+

N2H4 + H2O <--> N2H5+ + OH- this is Kb

so

Ka = Kw/KB = (10^-14)/(8.5*10^-7) = 1.176470*10^-8

C)

claculate hydrazine if pH = 8.5; pOH = 5.5

[OH-] = 10^-5.5 = 3.162*10^-6

N2H4 + H2O <--> N2H5+ + OH-

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

[N2H4] = (1.2*10^-2 - 3.162*10^-6)= 0.01199683M

d)

(8.5×107) = x * (3.162*10^-6) / (1.2*10^-2 - 3.162*10^-6)

x = (8.5*10^-7)/(3.162*10^-6) * (1.2*10^-2 - 3.162*10^-6)

[N2H5+] = 0.003224 M

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