The Kb for trimethylamine, (CH3)3N is 6.40x10-5. If 100 mL of 0.655 M trimethylamine
is titrated with 0.500 M HCl calculate the pH of the solution…
a) after 30.0 mL of HCl has been added to the solution. [10.3]
b) at the equivalence point. [5.18]
The [10.3] and [5.18] are the answers to the respective questions, but I am not sure how to get to this answer.
Follow the same steps as the NH4^+ earlier. It's worked the same way.The equation is (CH3)3N + HCl --> (CH3)3NH^+ + Cl^- For the equivalence point, the pH is determined by the hydrolysis of the salt. And this is worked the same way as the post you added to the NH4^+ earlier. I didn't show you how to do that but the equivalence point for this one is done this way.
At the equivalence point the hydrolysis of the salt, which is (CH3)3NH^+, is
.................(CH3)3NH^+ + HOH ==> (CH3)3N + H3O^+
You had 65.5 millimoles trimethyl amine to start and you added 131 mL of the 0.500 M HCl in the titration to the equivalence point so the total volume is 131 + 100 = 231 mL so the concentration of the salt is 65.5millimoles/231 mL = 0.284 M. Plug that into the hydrolysis equation like this and proceed.
.................(CH3)3NH^+ + HOH ==> (CH3)3N + H3O^+
I.................0.284..................................0................0
C..................-x......................................x................x
E.................0.284.................................x.................x
Ka for (CH3)3NH^+= (Kw/Kb for (CH3)3NH2) = [(CH3)3N][H3O^+]/[(CH3)3NH^+]. Plug in the numbers to get
(1E-14/6.40E-5) = (x)(x)/(0.284 - x). Solve for x and convert to pH.
I worked both problems and the answers you cited are correct.
To solve this problem, we will use the concept of acid-base titration and the equilibrium constant Kb for trimethylamine. Here's how to calculate the pH of the solution at different stages:
a) After 30.0 mL of HCl has been added to the solution:
Step 1: Calculate the moles of trimethylamine (CH3)3N initially present:
Moles of (CH3)3N = concentration x volume = 0.655 M x 0.100 L = 0.0655 moles
Step 2: Calculate the moles of HCl added:
Moles of HCl = concentration x volume = 0.500 M x 0.030 L = 0.015 moles
Step 3: Determine the limiting reactant:
Since the stoichiometry of the reaction between (CH3)3N and HCl is 1:1, the limiting reactant is (CH3)3N because it has fewer moles.
Step 4: Calculate the moles of (CH3)3N remaining after the reaction:
Moles of (CH3)3N remaining = initial moles - moles of HCl reacted
= 0.0655 moles - 0.015 moles
= 0.0505 moles
Step 5: Calculate the concentration of (CH3)3N after 30.0 mL of HCl has been added:
Concentration of (CH3)3N = moles remaining / volume = 0.0505 moles / 0.100 L
= 0.505 M
Step 6: Calculate the pOH of the (CH3)3N solution:
pOH = -log10(Kb)
= -log10(6.40x10^-5)
= 4.19
Step 7: Calculate the pH of the (CH3)3N solution:
pH = 14 - pOH
= 14 - 4.19
= 9.81
Therefore, the pH of the solution after 30.0 mL of HCl has been added is approximately 9.81.
b) At the equivalence point:
At the equivalence point, the moles of (CH3)3N will be completely neutralized by the moles of HCl.
Step 1: Calculate the total moles of HCl added at the equivalence point:
Moles of HCl = concentration x volume = 0.500 M x 0.030 L
= 0.015 moles (same as in part a)
Step 2: Calculate the moles of (CH3)3N initially present:
Moles of (CH3)3N = concentration x volume = 0.655 M x 0.100 L
= 0.0655 moles (same as in part a)
Since the stoichiometry of the reaction is 1:1, the moles of (CH3)3N remaining at the equivalence point will be zero.
Step 3: Calculate the concentration of (CH3)3N at the equivalence point (when (CH3)3N is completely reacted):
Concentration of (CH3)3N = moles remaining / volume = 0 moles / 0.100 L
= 0 M
Step 4: Calculate the pOH at the equivalence point:
pOH = -log10(Kb) = -log10(6.40x10^-5)
= 4.19 (same as in part a)
Step 5: Calculate the pH at the equivalence point:
pH = 14 - pOH
= 14 - 4.19
= 9.81
Therefore, the pH of the solution at the equivalence point is approximately 9.81.
To calculate the pH of the solution during the titration, we need to consider the reaction between trimethylamine ((CH3)3N) and HCl. Trimethylamine is a weak base, and HCl is a strong acid.
The balanced chemical equation for the reaction between trimethylamine and HCl is:
(CH3)3N + HCl → (CH3)3NH+ + Cl-
Given that Kb for trimethylamine is 6.40x10^(-5), we can set up an ICE (initial, change, equilibrium) table to determine the concentration of (CH3)3N and (CH3)3NH+ at each stage of the titration.
a) After 30.0 mL of HCl has been added:
Initially, the volume of trimethylamine is 100 mL, which can be converted to 0.10 L. The initial concentration of trimethylamine is 0.655 M, so we have 0.655 mol/L x 0.10 L = 0.0655 moles of (CH3)3N.
Since the reaction is 1:1, 0.0655 moles of (CH3)3N will react with 0.0655 moles of HCl. Since 30.0 mL (0.03 L) of HCl has been added, we have 0.500 mol/L x 0.03 L = 0.015 moles of Cl- ions.
Now, let's calculate the concentration of (CH3)3NH+ and (CH3)3N after the reaction:
(CH3)3N + HCl → (CH3)3NH+ + Cl-
0.0655 0.015
At equilibrium, the change in concentration will be equal, so:
(CH3)3N: 0.0655 - 0.015 = 0.0505 moles
(CH3)3NH+: 0 + 0.015 = 0.015 moles
Next, we need to calculate the concentrations of (CH3)3N and (CH3)3NH+ using the volume of the final solution, which is 100 mL + 30 mL = 130 mL = 0.130 L:
(CH3)3N: 0.0505 moles / 0.130 L = 0.3887 M
(CH3)3NH+: 0.015 moles / 0.130 L = 0.1154 M
To calculate the pH, we need to find the pOH and then convert it to pH. Since (CH3)3N is a weak base, we can assume that it mainly undergoes hydrolysis to form (CH3)3NH+ and OH- ions. The Kb expression is:
Kb = ([CH3)3NH+][OH-]) / [(CH3)3N]
We can assume that x is the concentration of OH- ions produced from the hydrolysis of (CH3)3N, and since OH- ions combine with H+ ions from the strong acid HCl to form water (H2O), the concentration of OH- ions is equal to the concentration of H+ ions produced from the reaction between HCl and (CH3)3N.
Therefore, [OH-] = [H+]. We can represent the concentration of H+ ions as x.
Kb = (x)(x) / (0.3887 - x) = 6.40x10^(-5)
As Kb is small, we can assume that x is small compared to 0.3887, so we can approximate 0.3887 - x as 0.3887.
6.40x10^(-5) = x^2 / 0.3887
Rearranging the equation and solving for x gives:
x^2 = 6.40x10^(-5) * 0.3887
x^2 = 2.485x10^(-5)
x = √(2.485x10^(-5))
x = 0.004985 (~0.005)
The concentration of H+ ions is approximately 0.005 M. Therefore, the pOH can be calculated as:
pOH = -log [OH-]
pOH = -log (0.005)
pOH ≈ 2.30
Finally, to calculate the pH, we can use the equation:
pH = 14 - pOH
pH = 14 - 2.30
pH ≈ 11.7
Therefore, the pH of the solution after 30.0 mL of HCl has been added is approximately 11.7.
b) At the equivalence point:
The equivalence point occurs when the moles of HCl added are equal to the moles of (CH3)3N initially present in the solution.
Since we have 0.0655 moles of (CH3)3N initially, the volume of HCl required to reach the equivalence point can be calculated using the balanced equation:
(CH3)3N + HCl → (CH3)3NH+ + Cl-
1 mol 1 mol
The moles of HCl needed to react with 0.0655 moles of (CH3)3N is also 0.0655 moles.
Using the concentration of HCl (0.500 M), we can calculate the volume of HCl needed to reach the equivalence point:
Volume of HCl = Moles of HCl / Concentration of HCl
Volume of HCl = 0.0655 moles / 0.500 M
Volume of HCl = 0.131 L = 131 mL
At the equivalence point, the solution will be a mixture of the conjugate acid (CH3)3NH+ and the anion Cl-. Since (CH3)3NH+ is the conjugate acid of a weak base, it will act as a weak acid in water.
We can consider (CH3)3NH+ as the only source of H+ ions. Therefore, the concentration of H+ ions can be calculated by dividing the moles of (CH3)3NH+ ions by the volume of the solution at the equivalence point:
Moles of (CH3)3NH+ = 0.0655 moles
Volume of solution = 100 mL + 131 mL = 231 mL = 0.231 L
Concentration of H+ ions = Moles of (CH3)3NH+ / Volume of solution
Concentration of H+ ions = 0.0655 moles / 0.231 L
Concentration of H+ ions ≈ 0.284 M
To calculate the pH, we can use the equation:
pH = -log[H+]
pH = -log(0.284)
pH ≈ 0.547
Therefore, at the equivalence point, the pH of the solution is approximately 5.18.