calculate the Ph of 0.3 M sodium propanoate, Ka 2.0 x 10 base -3. then calculate the pH after 0.03 mol NCL is added to 2L of the first solution.
Sodium propanoate we will call NaP. It's the propanoate ion that is hydrolyzed in solution.
.......P^- + HOH ==> HP + OH^-
I.....0.3............0....0
C......-x............x....x
E.....0.3-x..........x....x
Kb for P^- = (Kw/Ka for HP) = (x)(x)/(0.3-x)
Solve for x = (OH^-) and convert to pH.
For the second part, is NCl a typo for NaCl? Whether yes or no, write the reaction between HP and NCL (or whatever).