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?
To calculate the percent dissociation of a weak acid, you need to compare the concentration of the dissociated acid (in this case, the concentration of H+) with the initial concentration of the acid (HP).
The percent dissociation is given by the formula:
Percent Dissociation = ([H+] / initial concentration of HP) * 100
From your calculations, you obtained the concentration of H+ (formed from the dissociation of HP). Now, you need to divide the concentration of H+ by the initial concentration of HP and multiply by 100 to determine the percent dissociation.
For example, let's say you calculated the concentration of H+ to be 0.016 M, and the initial concentration of HP was 0.21 M:
Percent Dissociation = (0.016 M / 0.21 M) * 100 = 7.6%
So, in this case, the percent dissociation of the propanoic acid solution would be 7.6%.