00:01
All right, so we're told that the buffer is prepared.
00:03
We're given 0 .08 mole of propionic acid, that's propionic acid, and 0 .12 molar sodium propionate, and we're given the ka value.
00:14
So the question says we should calculate the ph value of the pure buffer.
00:19
So for a, the ph value of the pure buffer is the pka of the acid plus log the concentration of the base, that's the propionate ion, over the concentration of the acid.
00:47
And so the pka first is basically minus log ka, right, minus log ka of the acid, so minus log 1 .34 times 10 to the power of minus 5, and that will give you 4 .87, 4 .87.
01:14
So the ph of the solution is 4 .87 plus the log the concentration of the base, log of 0 .12 divided by 0 .08.
01:32
And when you punch a calculator, you will have the answer to 5 .0 as the ph, right, 5 .0.
01:40
That's going to be the ph of the pure buffer.
01:43
Now the b part says, we should write the chemical equation of the reaction which will cause when small amount of the strong base, you know, is added, right.
01:56
So if you add a small base, what will happen is that the acid, the propionic acid, would react with the base, and then you would form the conjugate base of the weak acid and water, right.
02:15
That's what you're going to have.
02:16
So this is the equation, which would mean that the concentration of this will decrease while the concentration of this will increase.
02:24
That's what you expect to have...