00:01
Hi, so in this case we're going to deal with finding the ph for two different compounds.
00:07
The first one is a hydroxylamine hydrochloride solution.
00:10
So this is going to be ho nh3 cl-.
00:19
This is all going to have a positive charge.
00:22
In this case we can omit directly the cl - because that is just going to be an expectation anion, which is going to be in solution with water.
00:32
Of course this is going to be a weak acid even though we're giving the value for kb.
00:41
And this is going to produce ho nh2 cl - h3o.
00:58
So what we have here is the acid and the corresponding conjugate base.
01:04
But if you realize this reaction corresponds to a ka, right? because ka is going to be concentration of the base times concentration of h3o +, divided by the concentration of the acid that i had originally.
01:21
So how can i find the ka coming from kb? well, quite easy.
01:26
We can get ka times kb needs to be equal to kw.
01:31
So that means that ka is kw divided by kb.
01:36
In this case, substituting the values, remember that kw is 1 times 10 to the minus 14.
01:42
The value for kb for this compound is 1a times 10 to the minus 8, which is going to give us a ka of 9 .09 times 10 to the minus 7.
01:56
So the next thing that we can do is employ our ice table in order to get the final concentrations that we require.
02:07
So the initial concentration that i have is 0 .0403 molar.
02:13
The amount of change that i'm going to get, let's call it x with a negative sign.
02:19
So i take a libram, i'm going to get 0 .0403 minus x.
02:25
I'm going to produce plus x of both of these compounds.
02:28
So x and x.
02:29
And recall that ka is equal to concentration of ho and h2.
02:35
In this case, that is the conjugate base times concentration of h3o plus divided by the concentration of my original wake acid.
02:48
This becomes x squared divided by 0 .0403 minus x.
02:56
Now we know that the value for ka is very small, right? we have a value of 10 to the minus 7.
03:01
That means that x is going to be very small.
03:05
So we can approximate the approximation of 0 .0403 minus x.
03:13
It's going to be approximately 0 .0403.
03:18
This is going to be true if and only if x is lower or equal than the 5 % of this quantity, which we can verify at the very end of our calculations.
03:30
Using that approximation, this becomes x squared divided by 0 .0403.
03:37
Or, and in a way that you may have found somewhere else, x squared divided by c0, right? so with this, we can solve for x.
03:47
X is going to be equal to my ka, which is 9 .09 times 10 to the minus 7, multiplied by my initial concentration, 0 .0403.
04:00
And i'm going to take the square root of all of this.
04:04
This is going to give us 1 .91 times 10 to the minus 4.
04:18
And guess what? that x is the same as the concentration of h3o plus that we have in our solution.
04:26
Equal to concentration of h3o.
04:32
That means that the ph of the solution is going to be simply the logarithm, the minus logarithm of this concentration, right? minus logarithm of concentration of h3o plus, which turns out to be 3 .72 for this first solution.
04:51
So we're done with the first case...