What is the concentration of hydroxide ions in a solution of .31 M sodium acetate (NaCH3COO)? (The Kb for sodium acetat is 5.6 x 10^-10.)
Let's call acetic acid HAc and sodium acetate is NaAc.
The hydrolysis of Ac^- is
Ac^- + HOH ==> HAc + OH^-
Kb = (HAc)(OH^-)/(Ac^-) = 5.6E-10.
Set up x = HAc and x = OH^-; Ac = 0.31-x, substitute and solve for x.