C6H5COOH = HB
........HB ==> H^+ + B^-
I.......0.048...0....0
C........-x.....x....x
E.....0.048-x...x.....x
Ka = ? = (H^+)(B^-)/(HB)
You should find Ka in your text or notes. Substitute and solve for x = (H^+) and convert that to pH, pOH, and (OH^-).
What is [H+], [OH-], pH, and pOH for a 0.048mol/L solution of the weak benzoic acid, C6H5COOH?
1 answer
1 answer