# Chemistry, #5

posted by
**Taasha** on
.

For propanioic acid, HC3H5O2, Ka=1.3 x 10^-5, determine the concentration of the species present, the pH and the percent dissociation of a 0.21 M solution. Do this for H+, OH-, C3H5O2-, HC3H5O2, and the pH and percent dissociation.

If someone can please just help me get started!!

Lat's call propanoic acid, HP, to save time on typing. It's a weak acid so it ionizes partially as follows:

HP ==> H^+ + P^-

Initially HP = 0.21 M, (H^+) = 0 and (P^-) = 0.

For every x mole of HP that ionizes, (H^+) will be x and (P^-) will be x. The amount of HP remaining will be 0.2 - x.

Set up Ka.

Ka = (H^+)(P^-)/(HP) = 1.3 x 10^-5

Plug in x for (H^+) and (P^-) and 0.21 - x for (HP). solve for (H^+) and pH. (P^-) = (H^+) and (HP) = 0.21 - x. I will let you finish. Post your work if you get stuck.

I figured out the concentrations, with your help! (thank you), but I am not quite sure how to figure out the percent dissociation. How do I go about this?