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.
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
2 answers
(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
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
0 answers