00:04
To calculate the ph of a 1 .2510 of the negative 2 molar solution of the decongestant of fendron -hardicloride, we're going to start by starting with the pkb is given us 3 .86.
00:17
So let's turn this into kb, which is 10 to the minus 3 .86, which will give us 1 .4 times 10 to the negative 4.
00:35
Fetrine hcl dissolved in water produces an acidic solution, therefore we need the ka, which is kw over kb, kw over the conjugate kb, so 1 .0 exponent 14 minus divided by 1 .4 exponent 5 minus.
00:56
And we will get 7 .1 times 10 to the negative 10.
01:12
Fedrin, hydrochloride will produce fedrine plus h plus.
01:24
Let's create our ice table here.
01:27
We were told that this is 0 .125 as the initial 0 and 0 change equilibrium minus x plus x plus x.
01:37
0 .125 minus x, x, x and x.
01:41
Let's plug this into our k -a expression, which will be 7 .1 times 10 and negative 10.
01:53
I'm going to double check this here.
02:12
Sorry, i made a mistake over here...