starting with 15.0 ml of 0.370M of organic base pyridine C5H5N a titration is carried out using 0.100M HCL. how do i find the pH before titration, after adding 22.0mL of the acid , at equivalent point and after 41.0mL is of acid is added? where do i begin?
at the beginning you have a "pure" solution of pyridine with Kb = ?
........BN + HOH ==> BNH^+ + OH^-
I....0.370M...........0......0
C.......-x............x......x
E....0.370-x..........x......x
Kb = etc. Substitute into the Kb expression and solve for x = (OH^-), then convert to pH.
After 22.0 mL of 0.1M HCl.
You have 15.0mL x 0.370M BN = M x L = mols.
You have added M x L HCl = mols.
.........BN + HCl ==> BNCl
Fill in the ICE chart, see how much of the BN is left and how much of the salt (BNCl) is formed and use the Henderson-Hasselbalch equation (that's a buffered solution you have made since you have a weak base and its salt present) and calculate pH.
At the equivalence point.
Determine where the equivalence point is.
mLacid x M acid = mL base x M base
Then recognize that the pH at the equivalence point is simply the pH of the salt due to hydrolysis.
(BN) = mols BN/L = estimated 5.5 mmols/about 60 mL = about 0.09M
..........BNH^+ + H2O ==> H3O^+ + BN
I.........0.09..............0.....0
C.........-x................x......x
E.......0.09-x.............x.......x
Ka for BNH^+ = (Kw/Kb for BN) = (x)(x)/(0.09-x)
Solve for x = (H3O^+) and convert to pH.
After 41.0 mL acid is added.\
Set up the same ICE chart as in the 22.0 mL EXCEPT this time you will notice that you are PAST the equivalence point so you have a solution of OH^-. Convert to pOH and to pH.
Post your work if you get stuck.