Question

Calculate the pH of a solution when 23.00 mL of 0.27 M Acetate is titrated with 19.35 mL of 0.25 M HCl. (Kaacetic acid = 1.75 x 10-5)

Answer 1

millimols acetate (Ac^-) = mL x M = 23 x 0.27 = 6.21
millimols HCl = 19.35 x 0.25 = about 4.84.

..................Ac^- + H^+ ==> HAc
I.................6.21......0.............0..........
add......................4.84..........................
C.............-4.84...-4.84..........+4.84
E.............abt 1.37...0.............4.84
(Ac^-) = about millimols/mL = 1.37/(23+19.35) = ?
(HAc) = about 4.84/(23+19.35) = ?
Plug into the Henderson-Hasselbalch equation and solve for pH.
Post your work if you get stuck.