a flask containing 50ml of 0.128M sodium acetate is titrated with 0.138M nitric acid. Calculate the initial ph of the acetate solution and calculate the ph after the addition of 24.9ml of nitric acid. pKa for acetate 4.757

Answer 1

a. I will call acetate Ac^-.
......Ac^- + HOH ==> HAc ==> OH^-
I...0.128.............0......0
C......-x.............x......x
E...0.128-x...........x......x

Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.128-x)
Solve for x = (OH^-) and convert to pH. I let mmol sand for millimols.
mmols Ac^ initially = mL x M = ?
mmols HNO3 added = mL x M = ?
The HNO3 neutralizes some of the OH^- and that forces the Ac^- to hydrolyze further.
mmols Ac^- remaining = mmols to start - mmols HNO3 added. This is the initial concn in the next step.

For the next step, H^+ from the HNO3 adds to the Ac^- so you have an acetate/acetic acid buffer.
Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log (base)/(acid)
........Ac^- + HNO3 ==> HAc
I.......?.......0........?
etc.

Post your work if you get stuck.