Question

1). In a titration 20 mL of 0.100 M of HCl is added to 50.0 mL of 0.150 M of ammonia.
A). What is the pH of NH3?

B). What is the pH of the resulting solution?

HELP :(

Answer 1

1) Makes no sense to me. Is that before the HCl is added or after. If after it's the pH of the solution, not NH3.

2)
HCl + NH3 ==> NH4Cl
mols HCl = M x L = 0.002
mols NH3 = M x L = 0.0075.
So you have how much NH3 that doesn't react? That's 0.0075 - 0.002 = 0.0055 mols NH3 excess. That's in how much volume? That's 50 mL + 20 mL = 70 mL = 0.07L so the excess NH3 concn is
mols/L = 0.0055/0.07 = 0.786M.
.......NH3 + H2O ==> NH4^+ + OH^-
I.....0.0786..........0........0
C......-x.............x........x
E.....0.0786-x........x........x

Substitute the E line into Kb expression for NH3 and solve for x = (OH^-), convert to pH.
Round to the correct number of significant figures.