Question

If a 0.10 M solution of an acid has a pH of 4.5, what would be the predicted pH of a .10 M solution of the conjugate, and why?

Answer 1

Let's call the unknown acid HA.
pH = -log(H^+)
-4.5 = log(H^+)
(H^+) = about 3E-5 but you need to do it more accurately.
...........HA ==> H^+ + A^-
I.........0.1.....0......0
C..........-x.....x......x
E........0.1-x....x......x

So x = about 3E-5
Substitute into Ka expression and solve for Ka.
Ka = (H^+)(A^-)/(HA)
I get Ka estimated 1E-8

Then the hydrolysis of the conjugate base.
.......A^- + HOH ==> HA + OH^-
I......0.1............0....0
C......-x.............x....x
E......0.1-x..........x.....x

Kb for A^- = (Kw/Ka for HA) = (x)(x)/(0.1-x) and solve for x = OH^-, convert that to pOH then to pH.