00:01
Alright, so we're given the ph of a 0 .1 molar solution of propanoic acid, a weak organic acid, and we're asked to calculate the acid dissociation constant of propanoic acid.
00:15
So first we're going to start by writing the dissociation equation for propanoic acid, propanoic ion, and the hydrogen and of course the ka for this is going to be the concentration of the conjugate base, the acid, and divided by the concentration of the equilibrium concentration of the weak acid itself.
01:02
Now, since we're given the ph, right, the ph is negative log of the hydrogen ion concentration at equilibrium, which means that the hydrogen ion concentration would be the hydrogen ion concentration at equilibrium would be equal to 10 to the power of minus the ph which was 2 .93, and that would give us 10 to the power of minus 2 .93 would be 0 .00117, right, that's what we're going to have, 0 .00117 molar.
02:03
And definitely, right, when we go back to the dissociation equation, we'll say yes, ch3coch2cooh, set up our ice table, ch3coo, my, sorry, ch3ch2coo - and hydrogen ion.
02:35
Now the initial concentration of the acid was 0 .1, so 0 .1 molar.
02:43
None of this were present at the beginning.
02:45
After a while, some of this would have been lost, this would have been gained, that would have been gained, and at equilibrium, we would have had 0 .1 minus x, this would have been x, and this would have been x...