Question

The pH of 0.05 M benzoic acid is 2.24. Calculate the change in pH when 0.76 g of C₆H₅COONa is added to 18 mL of 0.50 M benzoic acid, C₆H₅COOH. Ignore any changes in volume. The K_avalue for C₆H₅COOH is 6.5 x 10^-5.

Answer 1

no of moles of C₆H₅COONa = W/G.M.Wt     = 0.76/144 = 0.0053moles

no of moles of C6H5COOH   = molarity * volume in L

                                             = 0.5*0.018 = 0.009 moles

PKa   = -logKa

         = -log6.5*10^-5

         = 4.187

PH = PKa + log[ C₆H₅COONa]/[C6H5COOH]

       = 4.187+ log0.0053/0.009

      = 4.187-0.2299 = 3.9571

Change in PH = 3.9571-2.24 = 1.7171