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...

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

Asked on by max170892

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jerichorayel | College Teacher | (Level 2) Senior Educator

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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%

 

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