Question

Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 0.000003) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO3 have been added.

a. 0.0 mL

pH =

b. 20.0 mL

pH =

c. 25.0 mL

pH =

d. 40.0 mL

pH =

e. 50.0 mL

pH =

f. 100.0 mL

pH =

Answer 1

The question involves calculating pH values during the titration of hydrazine with nitric acid at various titrant volumes, using stoichiometry and the dissociation constant of hydrazine.

Step-by-step explanation:

The student is asking for the calculation of pH at various points during the titration of a weak base, hydrazine (H2NNH2), with a strong acid, nitric acid (HNO3). Titration involves the gradual addition of one solution of known concentration (titrant) to another substance to reach the equivalence point, where the solution is neutralized. To calculate the pH, we use the provided concentrations and volumes, the dissociation constant of hydrazine, and the stoichiometry of the titration reaction.

Answer 2

A = 11.74
B = 9.18
C = 10.4
D = 8.10
E = 7.21
F = 1.70

For more context, you may check the explanation :)

Step-by-step explanation:

This is a basic titration problem where an amine, H2NNH2, is being titrated with an acid, HNO3. The reaction between the two is:

H2NNH2 + HNO3 → H2NNH3+NO3-

The Kb value for H2NNH2 is given as 0.000003, which allows us to calculate the Kb expression:

Kb = [H2NNH3+][OH-] / [H2NNH2]

At the start of the titration, before any HNO3 is added, we have only H2NNH2 in solution and no H2NNH3+ or OH-. Therefore, at the start of the titration:

Kb = [H2NNH3+][OH-] / [H2NNH2]

0.000003 = (x)(x) / (0.100)

x = 0.0055 M

So at the start of the titration, [H2NNH2] = 0.100 M and [OH-] = 0.0055 M. To find the pH, we can use the fact that:

pH + pOH = 14

a. Before any HNO3 is added, the [OH-] is 0.0055 M. Therefore:

pOH = -log(0.0055) = 2.26

pH = 14 - 2.26 = 11.74

The pH of the solution is 11.74.

b. At 20.0 mL of HNO3 added, we can calculate the moles of HNO3 added:

moles of HNO3 = (0.200 M)(0.020 L) = 0.004 mol

This amount of HNO3 reacts completely with the same amount of H2NNH2, so the moles of H2NNH2 remaining is:

moles of H2NNH2 = 0.100 mol - 0.004 mol = 0.096 mol

The total volume of the solution is now 0.100 L + 0.020 L = 0.120 L. Therefore, the concentration of H2NNH2 is:

[H2NNH2] = 0.096 mol / 0.120 L = 0.800 M

Using the Kb expression, we can find the [OH-]:

Kb = [H2NNH3+][OH-] / [H2NNH2]

0.000003 = (x)(0.004) / (0.800)

x = 0.000015 M

Therefore, the pOH is:

pOH = -log(0.000015) = 4.82

And the pH is:

pH = 14 - 4.82 = 9.18

The pH of the solution is 9.18.

c. At 25.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.625 M, the [OH-] is 0.00004 M, and the pH is 10.4.

d. At 40.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.200 M, the [OH-] is 0.00080 M, and the pH is 8.10.

e. At 50.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.100 M, the [OH-] is 0.00155 M, and the pH is 7.21.

f. At 100.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.0 M, the [OH-] is 0.02000 M, and the pH is 1.70.