calculate the PH at the 60ml of .200M aniline (kb=3.2x10^-6) with .200M HNO3 solution before acid is added after 0,19,70 ml of HNO3
To calculate the pH of a solution, we need to consider the dissociation of the species involved. In this case, aniline (C6H5NH2) reacts with nitric acid (HNO3).
First, we need to determine the limiting reagent, which is the reactant that gets completely consumed during the reaction. The reaction between aniline and HNO3 can be represented as:
C6H5NH2 + HNO3 → C6H5NH3+ + NO3-
Based on the given information, we have a 0.200 M solution of aniline and a 0.200 M solution of HNO3. The volume of the solutions is also provided.
Let's calculate the limiting reagent by comparing the number of moles of aniline and HNO3 present in their respective volumes:
Number of moles of aniline = concentration × volume = 0.200 M × 60 mL = 0.012 moles
Number of moles of HNO3 = concentration × volume
For 0 mL of HNO3 added:
Number of moles of HNO3 = 0.200 M × 0 mL = 0 moles
For 19 mL of HNO3 added:
Number of moles of HNO3 = 0.200 M × 19 mL = 0.0038 moles
For 70 mL of HNO3 added:
Number of moles of HNO3 = 0.200 M × 70 mL = 0.014 moles
Based on the stoichiometry of the reaction, we can see that aniline and HNO3 react in a 1:1 ratio. Therefore, the number of moles of aniline determines the extent of the reaction.
Now let's calculate the pH at the specified points:
1. For 0 mL of HNO3 added:
Since there is no HNO3, the reaction has not occurred, and the pH solely depends on the aniline. Aniline is a weak base, so we can use the Kb value (3.2x10^-6) to calculate the concentration of the aniline cation (C6H5NH3+). The equilibrium equation is:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
Using the Kb expression:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
As the concentration of OH- is determined by the concentration of C6H5NH3+:
[C6H5NH3+] = [OH-]
Substituting values:
3.2x10^-6 = [C6H5NH3+]^2 / [C6H5NH2]
[C6H5NH3+] = sqrt(Kb × [C6H5NH2])
[C6H5NH3+] = sqrt(3.2x10^-6 × 0.012 moles) ≈ 8.31x10^-4 M
Since the pH is calculated using the concentration of the hydronium ion (H3O+), we need to find [H3O+]. In a neutral solution, [H3O+] = [OH-] = 1x10^-7 M.
Using the equation for water dissociation:
[H3O+][OH-] = 1x10^-14
Substituting values:
[H3O+] × [C6H5NH3+] = 1x10^-14
[H3O+] × 8.31x10^-4 = 1x10^-14
[H3O+] ≈ 1.20x10^-11 M
To calculate pH:
pH = -log[H3O+]
pH = -log(1.20x10^-11)
pH ≈ 10.92
Therefore, at 0 mL of HNO3 added, the pH is approximately 10.92.
2. For 19 mL of HNO3 added:
The limiting reagent is aniline since we have a lesser number of moles. We need to determine how much aniline is left over after the reaction.
Amount of aniline reacted = initial moles of aniline - moles of aniline used
Amount of aniline reacted = 0.012 moles - 0.0038 moles = 0.0082 moles
Now, we need to find the concentration of the conjugate acid (C6H5NH3+) formed, which is responsible for the acidic nature of the solution. Since the reaction has gone to completion, the concentration of C6H5NH3+ is equal to the amount reacted.
[C6H5NH3+] = 0.0082 moles / 19 mL ≈ 0.432 M
Following the same procedure as above, we can calculate the pH of this solution. The pH will be lower than the pH at 0 mL of HNO3 added.
3. For 70 mL of HNO3 added:
The limiting reagent is now HNO3 since we have a greater number of moles. We need to determine the concentration of the excess HNO3 after the reaction.
Excess moles of HNO3 = total moles of HNO3 - moles of HNO3 used for the reaction
Excess moles of HNO3 = 0.014 moles - 0.012 moles = 0.002 moles
To calculate the concentration of the excess HNO3:
[Excess HNO3] = excess moles of HNO3 / 70 mL
[Excess HNO3] ≈ 0.002 moles / 70 mL ≈ 0.028 M
The pH of this solution will depend on the excess HNO3 concentration and can be calculated using the procedure mentioned above. The pH will be lower than both the previous cases since a strong acid is being added.
In summary:
- At 0 mL of HNO3 added, the pH is approximately 10.92.
- At 19 mL of HNO3 added, the pH will be lower than 10.92.
- At 70 mL of HNO3 added, the pH will be lower than both the previous cases.