A solution is prepared by mixing 37.5 mL of 0.20 M (CH3)3N with 50.0 mL of 0.20 M (CH3)3NHC and then diluting the mixture to a total volume of 100.0 mL.
Calculate the pH of resulting solution.
So far I have been able solve the mols of the two chemicals.
(CH3)3N being 0.20M/.375L = 0.53 mol
and (CH3)3NHCl as 0.20M/.5L = 0.4 mol
I was wondering if it was possible to calculate the pH without any k values (because in the original ques. I wasn't given any).
Now, according to the kb value of my textbook (6.5e-5) and going through the calculations I got a pH of 2.44
I am unsure if this is correct, because in this direction I couldn't seem to find a way to integrate the dilution.
Please help! I don't need the answer, just a way to solve this!
Use the Henderson-Hasselbalch equation.
You need pKa. If you have Kb, convert that to pKb (-log Kb) and then to pKa (pKa + pKb = pKw = 14). Plug this into the HH equation with EITHER M or mols and calculate pH. But note you didn't calculate mols.
mols = M x L = 0.20 x 0.375 = ?
mols = M x L = 0.20 x 0.050 = ?