Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 33 mL of 0.22 M C2H5NH2(aq) with 0.16 M HCl(aq). The Kb of ethylamine is 6.5 x 10-4

Answer 1

no of moles of C2H5NH2   = molarity*volume in L

                                       = 0.22*0.033   = 0.00726moles

at equivalence point

no of moles of C2H5NH2 = no of moles of HCl

0.00726                        = molarity * volume in L

0.00726                        =0.16* volume in L

volume in L   = 0.00726/0.16   = 0.045375L

------------C2H5NH2 + HCl ------------> C2H5NH3Cl

I -------- 0.00726 ----- 0.00726 --------------0

C------ -0.00726------ -0.00726--------------0.00726

E-------- 0 --------------- 0 --------------------- 0.00726

total volume in L = 0.045375 + 0.033 = 0.078375L

molarity of C2H5NH2Cl   = no of moles/total volume in L

                       = 0.00726/0.078375           = 0.0926 M

C2H5NH3Cl(aq) ----------> C2H5NH3^+ (aq) + Cl^-

0.0926M -------------------------0.0926M

---------------- C2H5NH3^+(aq) + H2O(l) ------------------> C2H5NH2 (aq) + H3O^+ (aq)

I -------------------0.0926 ----------------------------------------------0 -----------------------0

C------------------- -x ---------------------------------------------------+x --------------------- +x

E----------------- 0.0925-x --------------------------------------------- +x --------------------+x

               Ka = Kw/Kb

                       = 1*10^-14/(6.5*10^-4)   = 1.54*10^-11

              Ka   = [C2H5NH2][H3O^+]/[C2H5NH3^+]

            1.54*10^-11 = x*x/(0.0925-x)

            1.54*10^-11(0.0925-x) = x^2

              x   = 1.2*10^-6

    [H3O^+]   = x = 1.2*10^-6M

PH   = -log[H3O^+]

        = -log(1.2*10^-6)

       = 5.92 >>>>answer