At the beginning you have pure NH3 solution.
......NH3 + H2O ==> NH4^+ + OH-
I.....0.1............0.......0
C......-x............x.......x
E.....0.1-x...........x......x
Plug the E line into the Kb expression and solve for x = OH, then convert to pH.
For part b. You started with how many mols of each.
mols NH3 = M x L = ? about 0.005
mols HCl = M x L = ? about 0.002
So you have 0.003 M NH3 left no HCl in excess and you formed 0.002 mols NH4Cl.
Use the Henderson-Hasselbalch equation for buffers and calculate the pH.
pH = pKa + log (base)/(acid)
For the titration of 50.00mL of 0.1000 M ammonia with 0.1000 M HCl, calculate the pH
(a) before the addition of any HCl solution
(b) after 20.00mL of the acid has been added
2 answers
take it ez maaaan