Question

Sort the pH of the following mixtures from LOWEST (most acidic) to HIGHEST (most basic):

(I) 0.001 mol of HCl added to 1 L water

(II) 0.001 mol HCl added to a buffer solution of 0.1 M HCOOH and 0.1 M NaHCOO (pKa=3.75)

(III) 0.001 mol HCl added to a buffer solution of 0.1 M NH4Cl and NH3 (pKa=9.25)

Answer 1

I molarity of HCl = no ofmoles/volume of solution in L

                           = 0.001/1 = 0.001M

           HCl ----------------> H^+ (aq) + Cl^-(aq)

         0.001M                    0.001M

         [H^+]    = [HCl]

         [H^+]   = 0.001M

          PH    = -log[H^+]

                   = -log0.001   = 3

    PH = 3 more acidic

II. no of moles of HCOOH after addition of 0.001mole of HCL = 0.1+0.001   = 0.101moles

    no of moles of NaHCOO after addition of 0.001 moles of HCl = 0.1-0.001   = 0.099 moles

   PH = Pka + log[NaHCOO]/[HCOOH]

          = 3.75 + log0.099/0.101

            = 3.75-0.008686   = 3.7413

III)   no of moles of NH3 after addition of 0.001 moles of HCL = 0.1-0.001 = 0.099 moles

   no of moles of NH4Cl after addition of 0.001 moles of HCl   = 0.1+0.001 = 0.101 moles

    POH = Pkb + log[NH4Cl]/[NH3]

POH = 4.75 + log0.101/0.099

            = 4.75+0.008686   = 4.7586

PH   = 14-POH

          = 14-4.7586   = 9.2414 highest basic