Question

What is the original molarity of HF in a solution if the pH is 3.52 at equilibrium and the pKa is 9.61? This is all of the information we were given and i can’t seem to find the answer.

Answer 1

HF is weak acid

PH   = 1/2PKa -1/2logc

3.52 = 1/2*9.61-1/2logc

3.52 = 4.805 -1/2logc

-1/2logc = 3.52-4.805

-1/2logc   = -1.285

logc    = 2.57

logc   = 2.57

c = 10^2.57   

              = 371

second method

             PKa = 9.61

           -logKa   = 9.61

              Ka    = 10^-9.61   = 2.45*10^-10

             PH = 3.52

          -log[H^+] = 3.52

                [H^+]   = 10^-3.52   = 0.0003019

            HF ------------> H^+ + F^-

I          C                        0          0

C        -C                    C      C

E         C-C                 C        C

      Ka    = [H^+][F^-]/[HF]

      Ka    = C*C/C-C  

    Ka      = C^2/1-                                       [ <<<<<<1]

   Ka    = C^2

          = Ka/C

    [H^+]    = C

   [H^+]     = CKa/C

    [H^+]    = KaC

    0.0003019 = 2.45*10^-10*C

taking both sides squar

9.11*10^-8 = 2.45*10^-10 *c

    C                 = 9.11*10^-8/2.45*10^-10    = 371M

initial conc of HF = 371 >>>>>>answer