A 100.0-mL sample of 0.250 M aniline (C6H5NH2) is titrated with 0.500 M HCl. What is the pH at the
equivalence point?
The pH at the equivalence point is determined by the hydrolysis of the salt, aniline hydrochloride.
....C6H5NH3^+ + H2O ==> H3O^+ + C6H5NH2
I.....0.167M.............0........0
C........-x..............x........x
E.....0.167-x............x........x
Ka for the salt - (Kw/Kb for aniline)=(H3O^+)(C6H5NH2)/(0.167-x)
salt concn determined as follows:
100 mL x 0.250M = 25.0 millimoles.
M = mmol/mL = 25/150 = 0.1667M
Solve for x = (H3O^+) and convert to pH.
ьызИ
Well, at the equivalence point of aniline and HCl, it's like a game of tug of war between aniline and HCl! They reach a perfect balance, like two clowns on a seesaw. So, the pH at the equivalence point depends on which clown is the heavyweight - aniline or HCl!
In this case, aniline is a weak base and HCl is a strong acid. Now, imagine the aniline clown being a little bit stronger than the HCl clown. The pH at the equivalence point would be slightly basic because aniline would have a tiny edge in balancing out the acidity of HCl.
However, since aniline is a weak base, it won't completely neutralize the HCl. So, unfortunately, I can't give you an exact pH value. But just remember, at the equivalence point, it's a clowns' balancing act between aniline and HCl!
To find the pH at the equivalence point, we first need to determine the moles of acid and base present.
Step 1: Write the balanced chemical equation for the reaction between aniline (C6H5NH2) and HCl.
C6H5NH2 + HCl → C6H5NH3+ + Cl-
Step 2: Calculate the moles of aniline (C6H5NH2) present in the 100.0 mL of 0.250 M solution.
moles of aniline = volume (L) × concentration (mol/L)
moles of aniline = 0.100 L × 0.250 mol/L
moles of aniline = 0.025 mol
Since the stoichiometry between aniline and HCl is 1:1, this means we also have 0.025 mol of HCl.
Step 3: Calculate the total volume of the solution at the equivalence point.
To reach the equivalence point, we need equal moles of aniline and HCl. Since both have 0.025 mol, the total moles at the equivalence point will be 0.025 + 0.025 = 0.05 mol.
Step 4: Calculate the total volume of the solution at the equivalence point.
To find the total volume, we can use the dilution equation:
moles = concentration × volume
volume = moles / concentration
volume = 0.05 mol / 0.500 mol/L
volume = 0.100 L = 100 mL
So, the total volume of the solution at the equivalence point is 100 mL.
Step 5: Calculate the pH at the equivalence point.
At the equivalence point, equal amounts of aniline and HCl react to form a neutral salt, C6H5NH3Cl. This salt will dissolve in water to form C6H5NH3+ ions and Cl- ions. Since C6H5NH3+ is a weak base, it will undergo hydrolysis in water, and the solution will be slightly basic.
To calculate the pH at the equivalence point, we need to consider only the hydrolysis of C6H5NH3+. The hydrolysis equation is:
C6H5NH3+ + H2O ⇄ C6H5NH2 + H3O+
Since this is a 1:1 reaction, the concentration of C6H5NH3+ and H3O+ will be the same.
Let x be the concentration of H3O+.
Using the equilibrium constant expression, pH can be calculated as:
Kw = [H3O+][OH-]
Kw = (x)(x)
Kw = x^2
From the equation, we know that the equilibrium constant Kw is equal to 1.0 × 10^-14.
Solving for x:
1.0 × 10^-14 = x^2
x = sqrt(1.0 × 10^-14)
x = 1.0 × 10^-7 mol/L
So, the concentration of H3O+ is 1.0 × 10^-7 mol/L at the equivalence point.
Using the relationship between H3O+ concentration and pH, we can calculate the pH:
pH = -log[H3O+]
pH = -log(1.0 × 10^-7)
pH = -(-7)
pH = 7
Therefore, the pH of the solution at the equivalence point is 7.
To find the pH at the equivalence point of the titration, we first need to determine the stoichiometry of the reaction between aniline (C6H5NH2) and HCl. From the chemical equation:
C6H5NH2 + HCl -> C6H5NH3+Cl-
We can see that for every one mole of aniline reacted, one mole of HCl is consumed. This means that at the equivalence point, the moles of aniline and HCl will be equal.
First, let's calculate the number of moles of aniline in the 100.0 mL of 0.250 M aniline solution:
moles of aniline = volume (L) × concentration (M)
= 100.0 mL × 0.250 mol/L
= 0.0250 mol
Since the number of moles of aniline is equal to the number of moles of HCl at the equivalence point, we can calculate the volume of HCl needed to reach this point.
moles of HCl = moles of aniline
volume of HCl = moles of HCl / concentration of HCl
Using the given concentration of HCl (0.500 M), we can calculate the moles of HCl:
moles of HCl = 0.0250 mol
Using the equation for volume:
volume of HCl = 0.0250 mol / 0.500 mol/L
= 0.0500 L
= 50.0 mL
Therefore, at the equivalence point, 50.0 mL of 0.500 M HCl will be required to completely react with the 100.0 mL of 0.250 M aniline solution.
Now, let's consider the chemical reaction that occurs between aniline and HCl. Aniline is a weak base, while HCl is a strong acid. When a weak base reacts with a strong acid, a salt is formed along with water. In this case, the salt formed is C6H5NH3+Cl-, which is the conjugate acid of aniline.
At the equivalence point, the moles of the aniline salt C6H5NH3+Cl- will be equal to the moles of the original aniline, and no aniline or HCl will be left in the solution. The resulting solution will only contain the salt.
The salt C6H5NH3+Cl- is the ammonium ion (NH4+) combined with the benzenide ion (C6H5-) from aniline. Therefore, the solution will be acidic due to the presence of the NH4+ ion. The pH of the solution can be calculated by considering the dissociation of NH4+ in water:
NH4+ + H2O <-> H3O+ + NH3
This is the reaction of the ammonium ion with water, where H3O+ is the hydronium ion and NH3 is ammonia. In this case, the solution will be a buffer solution, as NH4+ is the conjugate acid of the weak base NH3 (ammonia).
To calculate the pH of the solution, we need to consider the equilibrium constant for the reaction:
Ka = [H3O+][NH3] / [NH4+]
At the equivalence point, the concentration of NH4+ (C6H5NH3+Cl-) is equal to the initial concentration of aniline, which is 0.250 M, and the concentrations of H3O+ and NH3 are unknown.
Since we have equal moles of NH4+ and NH3, we can assume that the concentrations of H3O+ and NH3 are the same.
Ka = [H3O+]² / [0.250]
[H3O+]² = Ka × [0.250]
[H3O+] = sqrt(Ka × [0.250])
Now, we need the pKa value of NH4+. The pKa value is the negative logarithm of the Ka value. The pKa of NH4+ can be found in a reference table or database, usually around 9.25.
Let's assume the pKa of NH4+ is 9.25. Now, we can calculate the pH of the solution at the equivalence point:
pH = 1/2 × (pKa - log([0.250]))
= 1/2 × (9.25 - log([0.250]))
Using a scientific calculator, we can substitute in the values and solve for pH:
pH = 1/2 × (9.25 - (-0.602))
= 1/2 × 9.852
pH = 4.926
Therefore, at the equivalence point of the titration of a 100.0 mL sample of 0.250 M aniline with 0.500 M HCl, the pH of the solution will be approximately 4.926.