At 25C the pH of an aqueous solution of propanoic acid (pKa=4.874) is measured at equilibrium to be 3.25. Answer the following questions   What is the equilibrium concentration of the conjugate base? Whats was the initial concentration of propanoic acid, if this Ph is measured? What is the ionization fraction of the propanioic acid? No initial concentrations are given

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Propanoic acid is a monoprotic weak acid. This means that at equilibrium, the reaction equation would be:

HA  <--->  H+  +  A-

HA = propanoic acid

A- = conjugate base

H+ = hydronium ions

We can get the value of H+ from the given pH which is 3.25.

pH =...

See
This Answer Now

Start your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your Subscription

Propanoic acid is a monoprotic weak acid. This means that at equilibrium, the reaction equation would be:

HA  <--->  H+  +  A-

HA = propanoic acid

A- = conjugate base

H+ = hydronium ions

We can get the value of H+ from the given pH which is 3.25.

pH = -log [H+]

3.25 = -log [H+]

[H+] = 10^(-3.25)

[H+] = 5.6234x10^-4

Next, we draw the ICE table. 

       HA           <--->      H+    +     A-

I       M                           0             0

C     -x                         +x           +x

E   M-x                           x             x

 

We can see that

x = [H+] = [A-] = 5.6234x10^-4 = 5.62x10^-4

M would be the initial concentration of the acid HA. To solve for that value, we will use the equilibrium expression.

ka = [H+][A-]/ M-X

ka = [x][x]/M-x

ka = 10^(-pka) = [x][x]/M-x

**remember that pka is given in the problem.

10^(-4.874) = (5.6234x10^-4)^2/ M-5.6234x10^-4

by algebra, M = 0.02422 = 0.024

 

% ionization = X/M *100

% ionization = (5.6234x10^-4/0.02422) x100

% ionization = 2.32%

 

Approved by eNotes Editorial Team