Question

A 5.55 −g sample of a weak acid with Ka=1.3×10−4 was combined with 5.20 mL of 6.20 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.35. What is the molar mass of the weak acid?

Answer 1

no of moles of NaOH = molarity * volume in L

                                   = 6.2*0.0052   = 0.03224moles

PKa = -logKa

         = -log(1.3*10^-4)

         = 3.886

------HA(aq)    + NaOH(aq) -----------> NaA(aq) + H2O(l)

----- x    ------    0.03224 ----------- 0.03224

---- -0.03224 --- -0.03224 ------- 0.03224

--- x-0.03224 --- 0.03224 -----   0.03224

[NaA]     = no of moles/volume in L

              = 0.03224/0.75

[HA]   =   no of moles/volume in L

           = x-0.03224/0.75

PH = 4.35

       PH    = Pka + log[NaA]/[HA]

      4.35    = 3.886 + log(0.03224/0.75)/(x-0.03224/0.75)

      4.35    = 3.886 + log(0.03224)/(x-0.03224)

     4.35-3.886   = log(0.03224)/(x-0.03224)

    0.464   = log(0.03224)/(x-0.03224)

    log(0.03224)/(x-0.03224)    = 0.464

     (0.03224)/(x-0.03224)     = 2.91

     0.03224   = 2.91*(x-0.03224)

        x   = 0.0433

no of moles of acid = 0.0433moles

molar    mass of acid     = mass of acid / no of moles

                                       = 5.55/0.0433   = 128.2g/mole >>>answer

molar mass of acid = 128.2g/mole