Calculate the pH of (1)0.150M KC2H3O2:

Question

(2)0.200M pyridine chloride, C5H5NHCl: and(3) 0.500M NaCl

Answer 1

I'll do the first one. The others are the same type although the reactions are different.
KC2H3O2--potassium acetate is a salt. The acetate ion is a base and it hydrolyzes in water.
C2H3O2^- + HOH ==>HC2H3O2 + OH^-
Set up an ICE chart and substitute into the equilibrium equation below as follows:
Kb = (Kw/Ka) = (HC2H3O2)(OH^-)/(C2H3O2^-)
You know Kw and Ka. (HC2H3O2) = x = (OH^-) and (C2H3O2^-) = 0.150-x
Solve for x which is (OH^-), convert to pOH and from there to pH.
(2)C5H5NHCl is a salt. The C5H5NH^+ is an acid.
C5H5NH^+ + H2O ==> H3O^+ + C5H5N

(3)NaCl is a salt. Neither the Na^+ nor the Cl^- hydrolyze; therefore, the pH is that of pure water.
H2O ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 1E-14