Question

Determine the pH of each of the following solutions.

0.21 M KCHO2
0.19 M CH3NH3I
0.19 M KI

I'm not sure how i should set up the ice chart, and what Ka I should be using for the solutions

Answer 1

For the first one.
This is potassium formate, a salt. The K ion is not hydrolyzed but the formate ion is in which it acts as a base.
CHO2^- + HOH ---> HCHO2 + OH^-

Kb = (Kw/Ka) = (OH^-)(HCHO2)/(CHO2^-)
So HCHO2 is x, OH^- is x, CHO2^- is 0.21-x, Kw you know and Ka is Ka for HCHO2 (formic acid). Solve for x which will be OH^-, convert to pOH then to pH.

The second one is a salt, also, and it is the cation that hydrolyzes as follows:
CH3NH3^+ + H2O ==> CH2NH2 + H3O^+ but it is worked the same way as the formate salt.
For the last one you must recognize that neither the K ion nor the I ion is ionized; therefore, the solution is just like table salt in water and the pH is determined by the ionization of water. pH = 7.0