Q.1 What will be the pH at the equivalence point during the titration of a 100 ml 0.2M solution of CH3COONa with 0.2M of solution of HCl?(Ka = 2*10^-5)
Q.2 Aniline behaves as a weak base.When 0.1M,50ml solution of aniline was mixed with 0.1M,25ml solution of HCl the pH of resulting solution was 8.Then calculate the pH of 0.001M solution of anilinum chloride.
To determine the pH at the equivalence point during the titration of a weak acid-strong base or weak base-strong acid, we need to understand the reaction that takes place at the equivalence point.
For question 1:
In the titration of a weak acid (CH3COOH) with a strong base (NaOH), the reaction is as follows:
CH3COOH + NaOH → CH3COONa + H2O
At the equivalence point, the moles of CH3COOH will react completely with the moles of NaOH. Since the volumes of the solutions are given, we can calculate the moles of CH3COOH and NaOH.
To determine the pH at the equivalence point, we need to find the concentration of the acetate ion (CH3COO-) in the solution. At the equivalence point, the concentration of the acetate ion will be equal to the initial concentration of the CH3COONa.
The number of moles of CH3COONa can be calculated by multiplying the initial concentration (0.2 M) with the volume of the solution (100 mL = 0.1 L).
Moles of CH3COONa = 0.2 M * 0.1 L = 0.02 moles
Since CH3COONa is a salt of a weak acid (CH3COOH), it will undergo hydrolysis in water and produce CH3COO- ions.
Since CH3COOH is a weak acid, it will partially dissociate in water to produce CH3COO- and H+ ions. The dissociation of CH3COOH can be represented as follows:
CH3COOH ⇌ CH3COO- + H+
To determine the pH, we need to calculate the concentration of H+ ions. For this, we can make use of the Ka value, which is the equilibrium constant for the dissociation of CH3COOH.
The Ka expression for CH3COOH is:
Ka = [CH3COO-][H+] / [CH3COOH]
Since the concentration of CH3COO- and H+ ions are equal (as per stoichiometry), we can rewrite the expression as:
Ka = [H+]^2 / [CH3COOH]
Rearranging the equation, we can find the concentration of H+ ions:
[H+] = √(Ka * [CH3COOH])
Now, substitute the given Ka value (2*10^-5) and concentration of CH3COOH (0.2 M) into the equation to calculate the concentration of H+ ions:
[H+] = √(2*10^-5 * 0.2 M)
Finally, take the negative logarithm of the concentration of H+ ions to find the pH at the equivalence point:
pH = -log[H+]
For question 2:
To calculate the pH of the resulting solution when 0.1 M, 50 mL solution of aniline (a weak base) is mixed with 0.1 M, 25 mL solution of HCl (a strong acid), we need to understand the reaction that takes place between aniline and HCl.
The reaction between aniline and HCl can be represented as follows:
C6H5NH2 + HCl → C6H5NH3+Cl-
At the equivalence point, all the moles of aniline will react completely with the moles of HCl. Since the volumes of the solutions are given, we can calculate the moles of aniline and HCl.
To determine the pH of the resulting solution, we need to find the concentration of the anilinium ion (C6H5NH3+) in the solution. At the equivalence point, the concentration of the anilinium ion will be equal to the initial concentration of the aniline.
The number of moles of aniline can be calculated by multiplying the initial concentration (0.1 M) with the volume of the solution (50 mL = 0.05 L).
Moles of aniline = 0.1 M * 0.05 L = 0.005 moles
Since aniline is a weak base, it will partially react with HCl and produce the anilinium ion (C6H5NH3+) and Cl- ions.
Since aniline is a weak base, it will partially react with HCl to form the anilinium ion (C6H5NH3+) and Cl- ions. The reaction can be represented as follows:
C6H5NH2 + HCl ⇌ C6H5NH3+Cl-
To determine the pH, we need to calculate the concentration of H+ ions. Since the concentration of H+ ions can be assumed to be the same as the concentration of the anilinium ion at the equivalence point, we can use the concentration of the anilinium ion to calculate the pH.
Finally, take the negative logarithm of the concentration of H+ ions (which is equal to the concentration of the anilinium ion) to find the pH at the equivalence point:
pH = -log[anilinium ion]
Q.1 To determine the pH at the equivalence point during the titration of CH3COONa with HCl, we can use the Henderson-Hasselbalch equation.
First, let's calculate the initial moles of CH3COONa and HCl:
Number of moles of CH3COONa = concentration × volume
= 0.2 M × 0.1 L
= 0.02 moles
Number of moles of HCl = concentration × volume
= 0.2 M × 0.1 L
= 0.02 moles
Since the ratio of CH3COONa and HCl is 1:1, the moles of CH3COONa is equal to the moles of HCl at the equivalence point.
Now, let's calculate the concentration of CH3COOH at the equivalence point by dividing the moles by the total volume:
Concentration of CH3COOH = (moles of CH3COONa)/(total volume)
= (0.02 moles)/(0.2 L + 0.1 L)
= 0.067 M
Next, we can use the Ka value to calculate the pKa:
Ka = [H+][CH3COO-]/[CH3COOH]
[H+] = [CH3COO-]
[H+]^2/[CH3COOH] = 2*10^-5
[H+]^2 = 2*10^-5 * 0.067
[H+]^2 = 1.34*10^-6
Taking the square root of both sides:
[H+] = √(1.34*10^-6)
[H+] ≈ 3.66 * 10^-4 M
Now, we can calculate the pH using the logarithmic equation:
pH = -log[H+]
pH ≈ -log(3.66 * 10^-4)
pH ≈ 3.44
Therefore, the pH at the equivalence point during the titration is approximately 3.44.
Q.2 To calculate the pH of the 0.001M solution of anilinum chloride, we first need to consider the reaction that occurs between aniline (C6H5NH2) and HCl:
C6H5NH2 + HCl ⟶ C6H5NH3+Cl-
Aniline acts as a weak base and reacts with HCl to form the anilinium chloride salt.
Now, let's calculate the concentration of anilinium chloride:
Concentration of C6H5NH3+Cl- = (moles of aniline)/(total volume)
= (0.1 M × 0.05 L)/(0.075 L)
= 0.0666 M
Since the solution contains C6H5NH3+ ions, we can use the Kb value to calculate the pKb:
Kb = [C6H5NH2][OH-]/[C6H5NH3+]
[C6H5NH2] = [C6H5NH3+]
[C6H5NH2]^2/[C6H5NH3+] = Kb
[C6H5NH2]^2 = Kb * 0.0666
[C6H5NH2] = √(Kb * 0.0666)
Now, let's find the Ka value using the pKa + pKb = 14 relationship:
pKa + pKb = 14
pKa = 14 – pKb
pKa ≈ 14 – (-log(Kb * 0.0666))
Now we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
pH ≈ pKa + log(1)
pH ≈ pKa
Therefore, the pH of the 0.001M solution of anilinum chloride is approximately equal to the pKa value calculated using the relationship mentioned above.
You are starting with M x L = 0.020 moles acetate. The equivalence point with 0.20 M HCl will be at 100 mL; therefore, you will have 0.020 moles acetic acid formed and it will be in a volume of 200 mL; therefore, the concn acetic acid will be 0.020/0.200. Set up an ICE chart and solve for H^+, then convert to pH.
2. I assume the information in the first part is to calculate pKa for aniline but I am not positive of that. phNH2 is aniline.
phNH2 + HCl ==> phNH3^+ + Cl^-
You started with 50 mL x 0.1 M aniline (5 mmoles) and added 25 mL of 0.1 M HCl (2.5 mmols). That will form 2.5 mmoles of the aniline hydrochloride salt and you will have 5.0-2.5 = 2.5 mmoles aniline left over and the pH of the buffered solution is 8. Use the Henderson-Hasselbalch equation to calculate pKa.
pH = pKa + log(base/acid)
8 = pKa + log (2.5/2.5)
8 = pKa.
phNH3^+ + HOH ==> phNH2 + H3O^+
Set up an ICE chart for hydrolyzing the salt, then substitute into Keq.
Ka = (H3O^+)(phNH2)/(phNH3^+)
I looked in a table of Kb values for aniline and found 3.97 x 10^-10 which is not pKa of 8 so I don't know which value you are supposed to use. If you don't use 8 I don't know why the firt part of q 2 was involved.