You have 20.00 mL of a 0.1M aqueous solution of the weak base (CH3)3N (Kb = 7.4x10^-5). This solution will be titrated with 0.100 M HCl.
How do you find the pH of this solution before any acid has been added.
To save typing I will call (CH3)3N--triethyl amine--RN
...........RN + HOH ==> RNH^+ +OH^-
initial....0.1............0....0
change.....-x............x......x
equil......0.1-x.........x......x
Kb = (RNH^+)(OH^-)/(RN)
Substitute from the ICE chart above into Ka expression and solve for x = (OH^-) and convert to pH.
To find the pH of the given solution before any acid has been added, we need to determine the concentration of hydroxide ions (OH-) in the solution. Since the weak base (CH3)3N ionizes in water to produce hydroxide ions, we can use the Kb value to calculate the concentration of OH-.
The ionization reaction of (CH3)3N in water can be represented as follows:
(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-
Given:
Volume of solution (V) = 20.00 mL = 0.02000 L (converted to liters)
Concentration of (CH3)3N solution (C) = 0.1 M
Kb value for (CH3)3N = 7.4x10^-5
First, we can calculate the concentration of (CH3)3N by multiplying the molarity by the volume of the solution:
Concentration (n) = C × V = 0.1 M × 0.02000 L = 0.0020 moles of (CH3)3N
From the balanced equation, we can see that 1 mole of (CH3)3N produces 1 mole of OH-. Therefore, the concentration of OH- is also 0.0020 moles.
Next, we can use the Kb expression to find the concentration of OH- ions:
Kb = [OH-][CH3)3NH+]/[(CH3)3N]
7.4x10^-5 = [OH-] × [CH3)3NH+]/[0.0020]
Since the reaction is in equilibrium, we can assume that the concentration of OH- is x, and the concentration of (CH3)3NH+ is also x. Then, the concentration of (CH3)3N will be n - x, where n is the original concentration of (CH3)3N.
Thus, the equilibrium expression can be written as:
7.4x10^-5 = x * x / (0.0020 - x)
By solving this equation, we can find the concentration of OH-. However, to calculate the pH, we need to calculate the concentration of H+ using the concentration of OH-. Since water is amphoteric, it can also act as an acid by producing H+ ions.
Since OH- is the conjugate base of water (H2O ⇌ H+ + OH-), we can assume that the concentration of H+ ions will be equal to the concentration of OH- ions. Therefore, the concentration of H+ is also 0.0020.
Now, we can use the equation pH = -log[H+] to find the pH of the solution:
pH = -log(0.0020) = -(-2.70) = 2.70
Thus, the pH of the given solution before any acid has been added is approximately 2.70.