1.) What must be the molarity of an aqueous solution of trimethylamine (CH3)3N
if it has a Ph of 11.04? Kb = 6.3*10^-5
if someone can show the mechanics of doing this, will be appreciated much.
Think ammonia, NH3.
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3). You've probably solved 100 problem where you know pH of a NH3 solution and you calculate (NH3).
(CH3)3N does EXACTLY the same thing.
(CH3)3N + HOH ==> (CH3)3NH^+ + OH^-
Write Kb expression, look up Kb, calculate OH from pH and solve for (CH3)3N.
Post your work if you get stuck.
To determine the molarity of an aqueous solution of trimethylamine ((CH3)3N), you can use the equation for the base dissociation constant (Kb). Here's how you can calculate it step by step:
Step 1: Write the balanced equation for the dissociation of trimethylamine in water:
(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-
Step 2: Write the expression for the base dissociation constant (Kb):
Kb = [ (CH3)3NH+ ] [ OH- ] / [ (CH3)3N ]
Step 3: Determine the concentration of hydroxide ions (OH-) in the solution. We know that pH = -log [H+], and since water is neutral, we can assume [H+] = [OH-]. Therefore, [OH-] = 10^(-pH).
Given that the pH of the solution is 11.04, [OH-] = 10^(-11.04).
Step 4: Set up the expression for Kb using the given information:
Kb = [ (CH3)3NH+ ] [ 10^(-11.04) ] / [ (CH3)3N ]
Step 5: Rearrange the equation to solve for the concentration of (CH3)3N:
[ (CH3)3N ] = [ (CH3)3NH+ ] [ 10^(-11.04) ] / Kb
Step 6: Finally, recall that molarity (M) is defined as moles of solute per liter of solution. We can assume a volume of 1 liter for this calculation. Therefore, the molarity of the solution of trimethylamine is equal to the concentration of (CH3)3N in moles per liter.
Note: Molarity (M) = mol/L
I hope this breakdown helps you understand the process of finding the molarity of the solution of trimethylamine.