A 25.0 mL sample of a 0.0600 M solution of aqueous trimethylamine is titrated with a 0.0750 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.
I already calculated pH of the solution after 10.0 and 30.0 mL were added, but I am having trouble finding the pH after 20.0 mL is added.Please help me, thank you.
To find the pH of the solution after 20.0 mL of acid has been added, we need to consider the reaction between the trimethylamine and the HCl.
The reaction between trimethylamine ((CH3)3N) and HCl is as follows:
(CH3)3N + HCl -> (CH3)3NH+ + Cl-
The trimethylamine acts as a base, accepting a proton (H+) from the acid (HCl) to form its conjugate acid, (CH3)3NH+.
To solve this problem, we can use the Henderson-Hasselbalch equation, which relates the pH, pKa (or pKb), and the ratio of concentrations of conjugate acid and base.
pH = pKa + log ([A-]/[HA])
Given:
- Volume of trimethylamine solution = 25.0 mL
- Concentration of trimethylamine solution = 0.0600 M
- Volume of HCl added = 20.0 mL
- Concentration of HCl solution = 0.0750 M
- pKb of (CH3)3N = 4.19
Before the addition of HCl (at the start of the titration):
- The concentration of trimethylamine ((CH3)3N) = 0.0600 M
- The concentration of its conjugate acid ((CH3)3NH+) = 0.0 M, as no acid added yet.
After the addition of 20.0 mL of HCl, we can calculate the new concentrations:
Step 1: Calculating the moles of (CH3)3N in the 25.0 mL solution.
moles of (CH3)3N = volume x concentration
moles of (CH3)3N = 25.0 mL x 0.0600 M
moles of (CH3)3N = 1.50 mmol
Step 2: Determining the limiting reagent and calculating the moles of HCl added.
As the stoichiometric ratio between (CH3)3N and HCl is 1:1, the moles of HCl added is also 1.50 mmol.
Step 3: Calculating the new volume and concentration of the solution after the addition of HCl.
- Volume of the resulting solution = initial volume of trimethylamine + volume of HCl added
Volume = 25.0 mL + 20.0 mL = 45.0 mL = 0.045 L
- Concentration = moles of (CH3)3N / Volume
Concentration = 1.50 mmol / 0.045 L
Concentration = 33.3 mM
Step 4: Calculating the concentration of the conjugate acid ((CH3)3NH+).
The concentration of (CH3)3NH+ is equal to the concentration of HCl added, as they react in a 1:1 ratio.
Concentration of (CH3)3NH+ = 0.0750 M x (20.0 mL / 45.0 mL)
Concentration of (CH3)3NH+ = 0.0333 M
Step 5: Using the Henderson-Hasselbalch equation to find the pH.
pH = pKa + log ([A-]/[HA])
pH = 4.19 + log (0.0333 M / 33.3 mM)
pH = 4.19 + log (0.001)
pH ≈ 4.19 + (-3)
pH ≈ 1.19
Therefore, the pH of the solution after adding 20.0 mL of HCl is approximately 1.19.