mols (CH3)3N = M x L = ?
mols HClO4 = M x L = ?
mol(CH3)3N in excess = initial mols - mols HClO4 and mols (CH3)3N/total volume which is 23.4 mL + 6.67 mL (substitute as L).
This is a weak base that react with H2O to give
...(CH3)3N + H2O==> (CH3)3NH^+ + OH^-
Set up and ICE chart, solve for x = OH^- and convert to pH.
A 23.4 mL sample of 0.300 M trimethylamine, (CH3)3N, is titrated with 0.399 M perchloric acid.
After adding 6.67 mL of perchloric acid, the pH is
1 answer
1 answer