00:01
In question 52, you were asked to casually determine whether or not the salt or compound was going to produce a solution in particular ph ranges.
00:16
Now you're being asked to actually calculate those phs.
00:20
So let's do these one at a time.
00:23
The first one is hno2.
00:26
You'll notice that hno2 has a fairly large k -a -value, k -a -value, 4 .5 times 10 to the negative 4, which is relatively large in comparison to the other weak acids found on table 16 .2.
00:43
Because it's relatively large, it's likely that it's going to dissociate more than 5%.
00:48
So you can try to calculate the ph using the simple little equation.
00:53
The hydronium ion concentration is the square root of the concentration of the weak acid multiplied by its ka value.
01:01
This then gives us a hydronium concentration, which we can take the negative log of to get the ph.
01:08
And we get 2 .17.
01:10
This isn't quite the answer in the back of the book.
01:13
And if we calculate just this value right here, the square root of 0 .1, multiplied by 4 .5 times 10 to the negative 4, we'll find out that it is about 6 .something percent of the 0 .1.
01:29
So it's a little bit more than 5 % dissociated.
01:33
So that means we're going to have to go back and recalculate using the quadratic formula.
01:42
When we use the quadratic formula, we set it up this way.
01:45
Ka equals the hydrogen concentration squared, divided by the concentration of the weak acid minus the hydrogenium concentration.
01:58
Solving for the quadratic, i won't do that here.
02:02
I'll just show you the answer.
02:04
Says 6 .49 times 10 to the negative 3.
02:08
And then we take the negative log of that and we get the ph in the back of the book of 2 .19.
02:16
Then for the next one, we have...
02:23
Yeah, this up here was about 6 .5 % or so.
02:31
This next one was ammonium chloride.
02:35
So ammonium has a k -a value.
02:38
The k -a value is fairly small, at 5 .6 times 10 to the negative 10.
02:44
So we're probably safe calculating the hydrogen concentration by taking the square root of the concentration of ammonium, which will be the concentration of ammonium chloride because it completely dissociates, multiplied by the k -a value.
03:00
This then is the hydrogenium concentration, which we take the negative log of to get ph, and we get 5 .13.
03:07
The next compound is sodium fluoride.
03:10
Sodium does nothing to affect the ph, but fluoride has a kb value that's very small at 1 .5 times 10 to the negative 11.
03:20
To calculate the ph will first solve for the hydroxide concentration.
03:26
The hydroxide concentration will be the square root of the concentration of the base, multiplied by the kb value for fluoride, which is 1 .5 times 10 to the negative 11.
03:39
This down here in the denominator is the hydroxide concentration.
03:44
If we divide that into kw, we get the hydronium concentration, and then we take the negative log of that to get the ph.
03:51
And we get ph 8 .09.
03:55
Next we'll solve for the ph of the 0 .1 molar magnesium acetate solution...