Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.0 10-6) 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
b)20.0 mL
c)25.0 mL
d)40.0 mL
e)50.0 mL
f)100.0 mL
I only need help with parts c) and d)
a)10.74
b)8.7
c)8.48
d)7.88
If you will show me what you've done so far and explain the parts you don't understand I can help you through those parts. Taken together, however, this is a BUNCH of typing.
To calculate the pH of the resulting solution after different volumes of HNO3 have been added, we need to consider the reaction that occurs during the titration. In this case, the reaction is between the weak base H2NNH2 and the strong acid HNO3. The reaction equation is as follows:
H2NNH2 + HNO3 -> NH4+ + NO2-
To determine the pH, we need to calculate the concentration of the resultant solution. Let's go step by step for each volume:
a) For 0.0 mL of HNO3 added:
No reaction has occurred yet, so the concentration of the solution remains the same. The pH can be calculated by considering the dissociation of H2NNH2 as a weak base. Since Kb is given, we know that H2NNH2 is a weak base with Kb = 3.0 x 10^-6.
Set up an ICE (Initial, Change, and Equilibrium) table to calculate the concentration of [OH-]. Initially, the concentration of H2NNH2 is 0.100 M, and there is no OH-:
H2NNH2 + H2O -> NH4+ + OH-
I 0.100 M 0 M 0 M 0 M
The equilibrium concentration of NH4+ will be the same as the initial concentration of H2NNH2, which is 0.100 M. The concentration of OH- will be x.
Then, use the Kb expression to set up an equation:
Kb = [NH4+][OH-] / [H2NNH2]
3.0 x 10^-6 = (0.100)(x) / (0.100)
Solving this equation, we find x = 3.0 x 10^-6 M. This represents the concentration of OH- ions produced by the weak base. To find the concentration of H+ ions, we use the equation Kw = [H+][OH-]:
1.0 x 10^-14 = (x)(3.0 x 10^-6)
Solving for x, we find x = 3.3 x 10^-9 M. This represents the concentration of H+ ions in the solution.
Finally, we can calculate the pH using the formula: pH = -log[H+]. Substituting the value of [H+], we find the pH to be 8.48.
b) For 20.0 mL of HNO3 added:
To find the concentration of the resulting solution, we need to find the moles of HNO3 added and calculate the new concentrations based on the reaction stoichiometry.
The initial moles of HNO3 are: (0.200 M)(0.020 L) = 0.004 mol.
The initial moles of H2NNH2 are: (0.100 M)(0.100 L) = 0.010 mol.
Since the acid and base react in a 1:1 ratio, the moles of HNO3 will be the limiting reactant and will be completely consumed. This means that the remaining amount of moles for H2NNH2 is 0.006 mol.
To find the new concentration of the solution, we divide the moles by the new volume (100 mL + 20 mL) = 0.120 L.
The new concentration of H2NNH2 is then: (0.006 mol) / (0.120 L) = 0.050 M.
We can proceed similar to part (a) to find the OH- concentration, H+ concentration, and pH.
c) For 25.0 mL of HNO3 added:
Follow the same steps as in part (b), considering the initial moles of HNO3 to be 0.200 M x 0.025 L.
d) For 40.0 mL of HNO3 added:
Follow the same steps as in part (b), considering the initial moles of HNO3 to be 0.200 M x 0.040 L.
e) For 50.0 mL of HNO3 added:
Follow the same steps as in part (b), considering the initial moles of HNO3 to be 0.200 M x 0.050 L.
f) For 100.0 mL of HNO3 added:
At this point, all of the H2NNH2 will be consumed, and we will have an excess of HNO3. In this case, the pH will be determined by the excess HNO3, which is a strong acid. Hence, the pH will be determined by the concentration of the acid.
Calculate the moles of HNO3 added: (0.200 M)(0.100 L) = 0.020 mol.
Calculate the new concentration: (0.020 mol) / (0.100 L + 0.100 L) = 0.100 M.
The pH can be directly calculated using the formula: pH = -log[H+]. Substituting the value of [H+], we find the pH to be 1.00.
To summarize, the pH values for each volume of HNO3 added are as follows:
a) 0.0 mL: pH = 8.48
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 = 1.00