00:01
Let's look at a buffer that contains 0 .15 molar propanoic acid and 0 .35 molar propanoate.
00:11
If you have something like sodium propanoate or propanoate with any alkali metal, it's going to completely dissociate in water into propanoate and sodium minus.
00:34
And since there's a one -to -one ratio of sodium.
00:38
Propanoid to propanoid, the concentration of sodium propanoid is always going to be equal to the concentration of propanoid itself.
00:48
So if you have 0 .35 molar sodium propanoid, that implies 0 .35 molar propanoid.
01:01
Let's look at it both.
01:02
Using these concentrations, we can solve for the ph and the hydrogenium ion concentration of the solution using the henderson -hauselbach equation.
01:30
So let's start by finding the p -k -a.
01:34
We have the k -a for proponoric acid, and we can go from k -a to p -k -a because we know that p -ka equals the negative log of the k -a.
01:48
So in this case the p -k -a is the negative log of 1 .3 times 10 to the negative 5th.
01:58
Let's look at a buffer 4 .8861...