Determination of the dissociation Constant (K_a) of a weak Acid. The pH of a 0.1

Question

Determination of the dissociation Constant (K_a) of a weak Acid. The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39, what is the K_a of the acid? K_a = Hydrofluoric Acid, A weak Acid with K_a = 7.1 times 10^-4 A. For a 0.50 M solution of HF, calculate the equilibrium concentrations of HF, H^+ and F^-. [HF] = [H^+] = [F^-] = % Dissociation =, pH = B. Now consider a 0.050 M solution of HF. In this case the error in neglecting x in the denominator is large and a quadratic equation has to be solved for x. Thus, using the quadratic equation solution, solve for [HF], [H^+], [F]. [HF] = [H^+] = [F^-] = % Dissociation =, pH =

Explanation / Answer

II.

pH of weakacid = 1/2(pka-logC)

c = concentration of weakacid = 0.1 M

pka = ?

   2.39 = 1/2(x-log0.1)

x = pka = 3.78

ka = 10^-pka

    ka = 10^-(3.78)

dissociation constant ( ka) = 1.66*10^-4

III.

A.

          HF(aq) <===> H^+(aq) + F^-(aq)

initial 0.5 M            0        0

at equil 0.5-x            x        x

ka = [H^+][F^-]/[HF]

7.1*10^-4 = x^2/(0.5-x)

x = [H^+]=[F^-] = 0.0185 M

[HF] = 0.5-0.018 = 0.482 M

pH = -log(H^+)

     = -log(0.0185)

    = 1.732

ka = cx^2

7.1*10^-4 = 0.5x^2

x = degree of dissociation = 0.0377

% of dissociation = 0.0377*100 = 3.77%

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