In Chemistry: what is ph during the titration of 20.00ml of 0.1000m triethylamine (ch3ch2)3n (kb=5.2x 10-4) with 0.1000m HCL solution after the additions of titrant : a) 0ml (b) 10.00ml (c) 19.00ml (d)15.00ml (e) 20.00ml (f) 25.00ml
The secret to these titration problems is to recognize where you are on the titration curve.
a)at the beginning of the titration you have pure triethylamine.
..........R3N + H2O ==> R3NH+^ + OH^-
initial...0.1M............0........0
change.....-x.............x.........x
equil....0.1-x............x..........x
Kb = (R3NH^+)(OH^-)/(R3N)
Substitute into Kb expression and solve for x, then convert to pH.
b), c), and d
mL x M = millimoles
mmoles R3N = 20.00 x 0.1000 = 2.00 mmols.
mmoles HCl = 10.00 x 0.1000 = 1.00 mmols.
...........R3N + HCl ==>R3NH^+ + Cl^-
initial....2.00
add...............1.00
change......-1.00..-1.00..1.00..1.00
equil.....1.00......0......1.00..1.00
Use Kb expression from a) and substitute as in this ICE table OR use the Henderson-Hasselbalch equation and solve for H^+ and/or pH.
c and d are worked the same as b.
e. At 20.00 mL acid you are at the equivalence point. The pH is determined by the hydrolysis of the salt. The salt will be 2.00 mmoles and the volume will be 40 mL; therefore, (R3NH^+) = 20/40 = 0.05M
........R3NH^+ + H2O ==> H3O^+ + R3N
initiial...0.05...........0.......0
change....-x..............x........x
equil....0.05-x............x........x
Ka = (Kw/Kb) = (H3O^+)(R3N)/(RcNH^+)
Substitute 1E-14 for Kw, Kb for the R3N which is listed in the problem, x and x for the numerator and 0.05-x for the denominator. Solve for x and convert to pH.
f)You have used all of the triethylamine (at 20.00 mL) so this last 5.00 mL is simply excess HCl. Since that is a strong acid, calculate the concn of HCl at that point and convert to pH.
(a) At 0 mL of HCl added, the solution only contains triethylamine and its conjugate acid, triethylammonium.
Since triethylamine is a weak base, it will react with water to form OH- and the conjugate acid, triethylammonium:
(CH3CH2)3N + H2O ⇌ (CH3CH2)3NH+ + OH-
The Kb expression for triethylamine can be used to calculate the OH- concentration:
Kb = [ (CH3CH2)3NH+ ][ OH- ] / [ (CH3CH2)3N ]
5.2×10^-4 = (x)(x) / (0.1000 - x)
Assuming that x is small compared to 0.1000, we can simplify the expression:
5.2×10^-4 = x^2 / 0.1000
x^2 = 5.2×10^-5
x = [OH-] = 7.21×10^-3 M
pOH = -log([OH-]) = -log(7.21×10^-3) = 2.14
pH = 14.00 - pOH = 14.00 - 2.14 = 11.86
Therefore, the pH of the solution before the addition of any HCl is 11.86.
(b) At 10.00 mL of HCl added, we can assume that the reaction between triethylamine and HCl is incomplete.
The amount of HCl added is 10.00 mL × 0.1000 M = 1.00×10^-3 mol.
The initial amount of triethylamine is 20.00 mL × 0.1000 M = 2.00×10^-3 mol.
The amount of HCl that has reacted is also 1.00×10^-3 mol.
Since HCl and triethylamine react in a 1:1 ratio, the remaining amount of triethylamine is (2.00×10^-3 mol - 1.00×10^-3 mol) = 1.00×10^-3 mol.
The total volume of the solution after the addition of 10.00 mL of HCl is 20.00 mL + 10.00 mL = 30.00 mL = 0.03000 L.
The concentration of triethylamine at this point is:
c = (1.00×10^-3 mol) / (0.03000 L) = 0.033 M
The concentration of the triethylammonium ion is also 0.033 M since the reaction is incomplete.
Using the Kb expression for triethylamine:
Kb = [ (CH3CH2)3NH+ ][ OH- ] / [ (CH3CH2)3N ]
5.2×10^-4 = (0.033 M)(x) / (0.033 M)
Solving for x gives:
x = [OH-] = 5.2×10^-4 M
pOH = -log([OH-]) = -log(5.2×10^-4) = 3.28
pH = 14.00 - pOH = 14.00 - 3.28 = 10.72
To find the pH during the titration, we need to calculate the concentration of OH- ions (which is the conjugate base of the triethylamine) at each addition of the titrant.
The reaction between triethylamine and HCl is as follows:
(CH3CH2)3N + HCl -> (CH3CH2)3NH+ + Cl-
(a) Initial Point (0 mL titrant):
At this point, no HCl has been added, so the concentration of OH- ions can be calculated using the Kb expression for the triethylamine:
Kb = [NH3+][OH-] / [NH2] = 5.2 x 10^-4
Since the initial concentration of triethylamine is 0.1000 M, the concentration of OH- ions can be calculated as follows:
5.2 x 10^-4 = (x)(x) / (0.1000 - x) ≈ x^2 / 0.1000
Solving the equation, we get x ≈ 0.0203 M (concentration of OH- ions). Now, we can calculate the pOH at this point:
pOH = -log10(0.0203) ≈ 1.69
Finally, we can find the pH using the equation:
pH = 14 - pOH = 14 - 1.69 ≈ 12.31
(b) After adding 10.00 mL of titrant:
The moles of HCl added can be calculated using the concentration and volume of the HCl solution:
moles HCl = 0.1000 M * 0.01000 L = 0.00100 mol
Since HCl and triethylamine are in a 1:1 stoichiometric ratio, the concentration of triethylamine will decrease by that amount:
0.00100 mol / 0.02000 L = 0.0500 M
Now, we can calculate the concentration of OH- ions using the new concentration of triethylamine:
Kb = [NH3+][OH-] / [NH2]
5.2 x 10^-4 = (x)(x) / (0.0500 - x) ≈ x^2 / 0.0500
Solving the equation, we get x ≈ 0.00662 M (concentration of OH- ions). Now, we can calculate the pOH at this point:
pOH = -log10(0.00662) ≈ 2.18
Finally, we can find the pH using the equation:
pH = 14 - pOH = 14 - 2.18 ≈ 11.82
(c) After adding 19.00 mL of titrant:
Using the same calculations as above, we would find that the concentration of OH- ions is approximately 0.00987 M. Hence, the pOH would be approximately 1.99, and the pH would be approximately 12.01.
(d) After adding 15.00 mL of titrant:
Using the same calculations as above, we would find that the concentration of OH- ions is approximately 0.00745 M. Hence, the pOH would be approximately 2.13, and the pH would be approximately 11.87.
(e) After adding 20.00 mL of titrant:
Using the same calculations as above, we would find that the concentration of OH- ions is approximately 0.0102 M. Hence, the pOH would be approximately 1.99, and the pH would be approximately 12.01.
(f) After adding 25.00 mL of titrant:
Using the same calculations as above, we would find that the concentration of OH- ions is approximately 0.0130 M. Hence, the pOH would be approximately 1.89, and the pH would be approximately 12.11.
So, the pH during the titration at each specified point would be approximately:
(a) 12.31
(b) 11.82
(c) 12.01
(d) 11.87
(e) 12.01
(f) 12.11
To determine the pH during the titration of triethylamine (CH3CH2)3N with HCl solution, we need to consider the acid-base reaction between the two.
The chemical equation for the reaction is:
(CH3CH2)3N + HCl → (CH3CH2)3NH+ + Cl-
Triethylamine is a weak base, and HCl is a strong acid. As the titration progresses, HCl is added to the triethylamine solution, neutralizing the base.
To find the pH at each addition point, we should compare the moles of HCl added with the moles of triethylamine originally present. We can use the concept of equilibrium to solve this problem.
Here's how to calculate the pH at each addition point:
a) 0 mL of titrant added:
At this point, no HCl has been added. The solution contains only triethylamine. Since triethylamine is a weak base, we assume that it barely dissociates in solution. Therefore, the concentration of triethylamine will remain unchanged. We calculate the pOH and then convert it to pH using the formula:
pOH = -log10(Kb)
where Kb is the base dissociation constant.
b) 10.00 mL of titrant added:
To determine the pH at this point, we need to determine the moles of HCl added. Since the concentration of the HCl solution is 0.1000 M, we can calculate the moles of HCl:
moles of HCl = volume (in L) × concentration (in mol/L)
Next, we need to determine the moles of triethylamine originally present. We know the volume (20.00 mL) and concentration (0.1000 M) of the triethylamine solution. Therefore:
moles of triethylamine = volume (in L) × concentration (in mol/L)
Now we can consider the stoichiometry of the reaction to determine the limiting reagent (either HCl or triethylamine) and calculate the excess moles. Finally, we find the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([salt]/[acid])
c) 19.00 mL of titrant added:
Follow the same steps as in part b) to determine the moles of HCl and triethylamine. Once again, consider the stoichiometry, calculate the excess moles of the limiting reagent, and find the pH using the Henderson-Hasselbalch equation.
d) 15.00 mL of titrant added:
Follow the same steps as in part b) to determine the moles of HCl and triethylamine. Consider the stoichiometry, calculate the excess moles of the limiting reagent, and find the pH using the Henderson-Hasselbalch equation.
e) 20.00 mL of titrant added:
At this point, the volumes of HCl and triethylamine are equal, suggesting a stoichiometric equivalence point. At this point, the pH can be determined by calculating the concentration of the final solution and applying the necessary formula.
f) 25.00 mL of titrant added:
To determine the pH at this point, calculate the moles of HCl and triethylamine, consider the stoichiometry, calculate the excess moles of the limiting reagent, and find the pH using the Henderson-Hasselbalch equation.
By following these steps, you should be able to determine the pH at each addition point during the titration of triethylamine with HCl solution.