Question

What is the pH and pOH for 100 ml of 1 M Acetic acid? What is the concentration in Molarity for 100 ml of CH3COON a solution that will get the pH of the same Acetic acid solution to 5? Ka = 1.8 * 10^-5

Answer 1

pka of aceticacid =-logka

                = -log(1.8*10^-5)

                = 4.745

pH of acetic acid = 1/2(pka-logC)

        pka = 4.745

         C   = concentration = 1 M

pH = 1/2(4.745-log1) = 2.3725

2) pH of CH3COONa = 7+1/2(pka+logC)

   2.3725 = 7+1/2(4.74+logC)

C = 1.011 M