Question

Consider the titration of a 23.4 −mL sample of 0.125 MRbOH with 0.110 M HCl. Determine each of the following.

Part A) the initial pH

Part B) the volume of added acid required to reach the equivalence point

Part C) the pH at 5.7 mL of added acid

Part D) the pH at the equivalence point

Part E) the pH after adding 6.0 mL of acid beyond the equivalence point

Answer 1

A. RbOH -------> Rb⁺ + OH^-

        0.125M                      0.125M

    [OH-]    = [RbOH]

   [OH-]   = 0.125M

   POH   = -log0.125

                = 0.9030

PH    = 14-POH

           = 14-0.9030   = 13.097

   HCl + RbOH --------> RbCl + H2O

no of moles of HCl = no of moles of RbOH

   M1V1                    = M2V2

0.11*V1                  = 0.125*23.4

         V1                   = 0.125*23.4/0.11    = 26.6ml

   M   = MBVB -MAVA/VA+VB

           = 0.125*23.4-0.11*5/23.4+5

           = 2.375/28.4   = 0.084M

[OH-] = M = 0.084M

POH   = -log[OH-]

            = -log0.084   = 1.0757

PH    = 14-POH

         = 14-1.0757   = 12.9243

D. M   = MAVA -MBVB/VA+VB
           =0.11*26.6- 0.125*23.4/23.4+26.6

           = 0.001/50   = 0.00002M

      [H+] = M = 0.00002M

     PH = -log0.00002

             = 4.6989