Posted by RZeal on .
Consider a 0.42 M solution of NaCH3COO
CH3COO–(aq) + H2O(l) -> CH3COOH(aq) + OH–(aq)
Calculate the pH of the solution at equilibrium. Try the weak acid or
weak base approximation (as appropriate) and check the validity of the
approximation. If the approximation is not valid, then solve the problem exactly.
Ka = Ch3COO- 5.6x10^-10
Ka = Ch3COOH 1.8x10^-5