Question

1)Calculate the pH during the titration of 20.00 mL of 0.1000 M HF(aq) with 0.2000 M LiOH(aq) after 4 mL of the base have been added. K_a of HF = 7.4 x 10^-4.

2)Calculate the pH during the titration of 20.00 mL of 0.1000 M morphine(aq) with 0.100 M HCl(aq) after 10 mL of the acid have been added. K_b of morphine = 1.6 x 10^-6

Thank you.

Answer 1

         HF + LiOH ---------> LiF + H2O

no of moles of HF = molarity * volume in L

                             = 0.1*0.02 = 0.002MOLES

no of moles of LiOH = molarity * volume in L

                                 = 0.2*0.004    = 0.0008moles

     Ka = 7.4*10^-4

   Pka = -logKa

           = -log7.4*10^-4

         = 3.1307

           HF + LiOH-------> LiF + H2O

    I     0.002     0.0008       0

   C     -0.0008 -0.0008     0.0008

E       0.0012   0              0.0008

PH   = PKa +log[LiF]/[HF]

         = 3.1307 + log0.0008/0.0012

        = 3.1307 -0.1760   = 2.9547 >>>> answer

2. no of moles of morphine = 0.1 *0.02 = 0.002moles

   no of moles of HCl           = 0.1*0.01 = 0.001moles

    Pkb   = -logKb

            = -log1.6*10^-6

            = 5.7958

   morphine + HCl -------> salt + H2O

      I         0.002 0.001    0

     C        -0.001 -0.001          0.001

    E         0.001         0                 0.001

    POH = PKb + log[salt]/[Base]

             = 5.7958 + log0.001/0.001

           = 5.7958 +0

     PH   = 14-PH

            = 14-5.7958 = 8.2042 >>>> answer