What is the pH of a .10M (CH3)3N solution? Kb=6.40e-5
To find the pH of the (CH3)3N solution, you need to understand that (CH3)3N is a weak base and its pH can be determined using the equilibrium constant expression for the base dissociation.
The dissociation reaction of (CH3)3N can be represented as follows:
(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-
The equilibrium constant expression for this reaction is:
Kb = [ (CH3)3NH+ ] [ OH- ] / [ (CH3)3N ]
Given that Kb = 6.40e-5 and the concentration of (CH3)3N is 0.10 M, we can assume the concentration of (CH3)3NH+ and OH- is initially zero (since (CH3)3N is a weak base).
Let x represent the concentration of (CH3)3NH+ and OH- ions formed at equilibrium. Since (CH3)3N initially has a concentration of 0.10 M, the equilibrium concentration of (CH3)3N will be (0.10 - x) M.
Using the Kb expression, we can write the equilibrium expression as:
6.40e-5 = x * x / (0.10 - x)
Now we need to solve this equation to find the value of x, which represents the concentration of OH-. We can use the quadratic formula:
x^2 = (6.40e-5)(0.10 - x)
Multiplying everything out, we get:
x^2 = 6.40e-5 * 0.10 - 6.40e-5 * x
Rearranging the equation, we have:
x^2 + 6.40e-5 * x - 6.40e-6 = 0
Now we can solve this quadratic equation. Once we have the value of x, we can calculate the OH- concentration, and then use that to find the pOH. Finally, we can convert pOH to pH using the equation:
pH = 14 - pOH
In this case, since we know the Kb value is relatively small, we can simplify the calculation and assume x is negligible compared to 0.10. This approximation allows us to calculate the OH- concentration and find the pH without solving a quadratic equation.
(CH3)3N + HOH ==> (CH3)3NH^+ + OH
Kb = [(CH3)3NH^+](OH^-)[CH3)3N]
Set up ICE chart, substitute into Kb expression above and solve for OH, convert to pH. Post your work if you get stuck.
I did my ice chart and found x=.00253
then i did the -log(.00253) to find the pH to be 2.597, but that's the wrong answer