Acids and bases
pH of a Weak Acid Calculating the pH of a weak acid
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- Sal: We now know all about strong acids.
- For example, hydrochloric acid in an aqueous solution.
- It's strong because it disassociate completely into
- hydrogen ions or hydronium molecules, depending on how
- you want to view it.
- And chloride anions.
- And it's an acid in the Arrhenius definition, because
- once again, this is an aqueous solution.
- In the Arrhenius definition because it is increasing the
- hydrogen concentration in the water.
- It's an acid in the Bronsted-Lowry definition
- because this thing is donating these hydrogen protons.
- And it's a Lewis acid because you can kind of view it as
- hydrochloric acid.
- By giving away this, it's taking these-- the chloride,
- as you can kind of say, it's taking the
- electrons for itself.
- So it's an electron acceptor.
- So that's another reason why it's a Lewis acid.
- But in every case, it completely disassociates.
- So it's a strong acid.
- Now, by implication, when I call this a strong acid, that
- must mean that there must be things called weak acids.
- And you could probably guess what they do.
- They don't disassociate completely.
- So let's say you have hydrofluoric acid in an
- aqueous solution.
- It does disassociate somewhat, so you are going to have some
- hydrogen break-off of the fluorine molecules.
- So a hydrogen ion, and that's of course
- in an aqueous solution.
- I was doing aqueous in blue.
- And then that's going to disassociate with the
- fluoride, the fluorine cation.
- But this is a weak acid, so it's
- not just in one direction.
- It actually goes in both.
- This is an equilibrium reaction.
- We know all about these now.
- Which means that some of this goes in that direction, some
- of that goes in that direction, and both of these
- species exist in the molecule.
- And just as aside, you might wonder hey, aren't fluorine
- and chlorine-- they're both halogens.
- Why is this a weak acid while this is a strong one?
- And this is just an aside.
- This is something I always used to wonder about.
- So in general, hydrogen likes to form bonds.
- Remember, alkali metals we included
- everything below hydrogen.
- Right?
- We included all of these guys.
- All of these guys, they have this 1 extra electron and
- they're on a valence shell, so they just want to give it up.
- And they tend to be very willing to give
- it up to these guys.
- So if you have any of these period 2 and below with a
- halogen, you normally form this ionic salt.
- Where this guy gives away an electron, and one of these
- guys takes it, and you have an ionic compound.
- Hydrogen's a little different.
- Hydrogen does have exactly 1 electron in
- it's outermost shell.
- It just has one shell.
- But the first shell is interesting because it's
- complete when it only has 2 electrons.
- These guys wouldn't have been happy.
- If you gave 1 electron to these guys-- they
- need to get to eight.
- So they're not happy.
- But if you give 1 electron to hydrogen,
- hydrogen is pretty happy.
- So when hydrogen forms bonds with these guys here, it
- doesn't just completely give up the electron.
- When hydrogen forms bonds with our halogens over here, it
- doesn't completely give up the electron.
- Because hydrogen says hey, if you give me some of the
- electron, then I get to 2.
- And then if you share the electron, you get to 8.
- So it does definitely like the bond with these guys, but
- doesn't give it completely up.
- Now.
- Hydrochloric acid, hydrobromic acid, and hydroiodic-- I think
- I think I'm seeing it right-- these are all strong acids.
- When hydrogen bonds with any of these guys.
- Which means it disassociates completely in water.
- But when hydrogen bonds with flourine, it doesn't
- disassociate completely in water.
- And the reason why is flourine is even more electronegative.
- So this bond between hydrogen and fluorine-- so if you
- compare hydrogen fluorine or hydrofluoric acid to hydrogen
- chloride, hydrogen fluoride to hydrogen chloride, this bond
- is much stronger.
- Not much stronger.
- I mean, you can measure in the energy of the bond, and we'll
- do that eventually.
- But this is a stronger bond than this right here.
- And hydrobromic is even weaker.
- Let me draw it.
- So this one is a little bit weaker than that one and then
- hydrobromic-- I'll draw it as one line right there.
- It's like that.
- So these are easier to break, so when you put this in water,
- hydrogen sees these water molecules that have these
- extra electron pairs floating around.
- And and it hogged electrons from here.
- So there's a partial negative charge, it's tempting for it
- to disassociate with the water molecules.
- Well, in this case, yeah, it's tempting, but it still like to
- be hanging around.
- Oxygen is very
- electronegative, but so is fluorine.
- So it likes to hang around with both.
- So it kind of doesn't completely disassociate.
- So that's why this one is a hydrofluoric acid, is a weak
- acid, while all of the other halogenic acids are strong.
- Well, with that said, the point of this isn't to just
- look at the periodic table.
- Let's figure out what the pH is when I dissolve-- let's say
- I have 1.5 molar of hydrofluoric acid.
- And obviously, by definition, since I've given this as
- molarity, I have it in some solutions.
- In this case, I have it in an aqueous solution.
- Now.
- You could look it up.
- Well actually, I'll just write it.
- So every equilibrium reaction has some equilibrium constant.
- So the equilibrium constant for this reaction-- I'll do it
- in green-- and since this is an acidic reaction, we'll say
- it's the equilibrium constant, or an acid constant, however
- you want to say it, for hydrogen fluorine.
- Hydrofluoric acid.
- Hydrofluoric acid is equal to the products.
- So it's the concentration of your hydrogen ions times your
- concentration of your fluorine ions divided by your
- concentration of your hydrogen fluoride, or
- hydrofluoric acid.
- So HFl.
- And if you look this up at 25 degrees Celsius at roughly
- room temperature-- I believe that is, but let's say it is--
- you get a constant value here of 7 times 10 to the minus 4.
- So if I have 1.5 molar of hydrofluoric acid, can I
- figure out what the pH is?
- Remember, esssentially, I want to solve for this right here.
- If I solve for this, then I know the hydrogen
- concentration.
- And then if I take the negative log of that, then I
- know its pH.
- So let's see if we can do that.
- So.
- I'm throwing 1.5 molar onto this side of
- the equation, right?
- But some of it's going to go and form that stuff, right?
- So let's say x molar gets disassociated.
- So on this side-- let me see if I can do this.
- So I'm throwing 1.5 molar on this side.
- Let's say x molar goes to that side, right?
- And so let's say I have 1.5 molar on this side.
- Once I'm in equilibrium, x goes to
- that side of the equation.
- Or x reacts to form this stuff, right here.
- Disassociates.
- So once I'm in equilibrium, I'll have my concentration of
- hydrogen fluoride will be 1.5 molar minus x.
- And what's my concentration of hydrogen?
- Well, if x molar, or x moles of this, the way you can view,
- disassociate, then we have x moles or x molar of that.
- And we'll also have x molar of that.
- Right?
- If one of the disassociates, then you get one of that and
- one of that.
- So these will be the equilibrium concentrations.
- And I think you can smell some algebra coming our way.
- So these are the equilibrium concentrations.
- Let's see if we can solve for x.
- So we know what the equilibrium constant is.
- It's 7 times 10 to the minus fourth.
- We know that the final equilibrium concentration of
- our hydrofluoric acid is the original 1.5 molar we put in,
- minus whatever disassociates into the stuff up here.
- And since x disassociated this stuff here, I have x molar of
- hydrogen, and I have molar of fluoride.
- This is just pure algebra now.
- Let's see if we can solve this.
- So this is equal to-- I'm going to
- need space to do this.
- So we get 7 times 10 to the minus 4.
- Let's multiply both sides times 1.5 minus x.
- Times 1.5 minus x.
- Is equal to
- x times x, is equal to x squared.
- And then if we multiply-- 7 times 10 to the minus fourth
- times-- 7 times 1.5, that's 10.5, right?
- So we have 10.5 times 10 to the minus fourth.
- That's that times that.
- Minus 7 times 10 to the minus fourth x
- is equal to x squared.
- This looks complicated, but it's just a quadratic
- equation, and there's many videos on that.
- So let's just throw all of this stuff onto to the
- right-hand side because I like having a
- positive coefficient here.
- So if I subtract this from both sides and add this to
- both sides, I get 0 is equal to x squared, plus this, so
- plus 7 times 10 to the minus fourth x, right?
- I just add this to both sides, so it shows up on this side of
- the equation, which is right here.
- Minus this.
- I'm subtracting this from both sides of the equation.
- So I get minus 10.5 times 10 to the minus 4.
- Now, this is a quadratic equation here.
- And actually, I'll throw you a little--
- We can solve this quadratic equation.
- We'll get the answer.
- In fact, I'm about to solve it.
- But there is an approximation you can do.
- Because if you look at this quadratic equation, this term
- here, x squared, dominates this term right here.
- Even when you have relatively small values of x, it
- dominates that term.
- So if chemists look at this and they say
- hey, you know what?
- If this term is so much smaller than what's going on
- here, in a lot of the cases, you might be able
- to just ignore it.
- And then the problem becomes very simple.
- Because if you ignore this term, it's still a quadratic,
- but you don't have to use a quadratic equation.
- Actually I'll do it on the side right here.
- I'll just do it in a box.
- Because this isn't the exact answer, but it's going to get
- it's pretty close to the exact answer.
- So in this-- I'll make a nice little box here.
- I'll make a box.
- So if we just ignored this term, then we'd have x squared
- is equal to this.
- Is equal to minus-- you would add this to the other side, so
- you would have plus 10.5 times 10 to the minus 4.
- Or x would be equal do the square root of this.
- x would be equal to-- what?
- See, 10.5-- I see you have to put the exponent thing here.
- So exponent for minus all right.
- And I want to take the square root, so to the 0.5 power is
- equal to 0.0324.
- So this is 0.0324.
- So if we make that simplifying assumption, we've essentially
- solved four our hydrogen concentration, right?
- That was x.
- This is x.
- So this is 0.0324.
- And if we want to figure out the pH, we just take the
- negative log of this.
- So let's do that.
- So you take the log of it.
- If you just press log, that's base 10.
- So you take the log.
- And we want the negative of that, so you just say minus
- 1.489, so about 1.49.
- So it's a pH of 1.49.
- But remember, I did that using my little simplifying
- assumption.
- It does not solve the quadratic equation.
- Now let's solve it, just to kind of confirm in our heads
- that we get the right answer.
- Because that's actually the right way to
- get the right answer.
- So quadratic equation.
- Minus b.
- So x is going to be equal to-- let me do it here.
- x is equal to minus b.
- So minus 7 times 10 to the minus 4 plus or minus the
- square root of b squared.
- So b squared is 49 times 10 to the minus 8, right?
- 7 squared.
- And then 10 to the minus 4 squared.
- Is minus 4 times 2 is minus 8.
- Minus 4 times a, which is 1, times c.
- Well, c is a minus sign there-- we have this minus
- sign, so it becomes plus.
- So 4 times 10.5 times 10 to the negative 4.
- That's what?
- That's 42 times 10 to the minus fourth power.
- All of that over 2a.
- a is just 1.
- All of that over 2.
- So let's see what we can get in this
- square root right here.
- And notice, this is a very small number compared to this.
- This is really all we should care about with the square
- root, but let's just do it.
- So 49 e 8 negative plus, so that's that small number--
- plus 42 times 10 to the negative-- or to the 4
- negative, is equal to that.
- So that's this value right here.
- And now we want to take the square root.
- So to the 0.5 power is equal to 0.06.
- So so far, we get x is equal to-- I will arbitrarily switch
- colors-- minus 7 times 10 to the minus 4
- plus or minus 0.0648.
- All of that over 2.
- And it should be clear.
- If we subtract it right here, we get a negative number.
- This is already a negative, so if we subrtact it, we get a
- negative numbers.
- So let's just worry about the plus sign because we can't
- have a negative concentration.
- So if we add that to this right there, so we have 0.06,
- and then we have minus--
- Actually let me make sure that the minus is there.
- So it's minus 7 times 10.
- Right.
- I just want to make sure I'm not making a mistake by
- putting a minus there.
- No.
- It looks right.
- So let's see.
- So we just do our math.
- So we take this and we subtract that.
- So it's minus-- I want to make sure you can
- see what I'm doing.
- So we have minus 7 e 4 negative is equal to that and
- then we have to divide by 2.
- Divided by 2 is equal to 0.032.
- So 0.032.
- And remember, this is our hydrogen concentration.
- It's also going to be our concentration of, of course,
- our fluorine ions.
- We don't care about that.
- We just want to figure out the pH.
- So this is 0.032, which is very close to our
- approximation before.
- But now we have the exact number.
- So let's take the log of it, and then take
- the minus of that.
- So the log of that.
- And then you want to put a minus in front of the log.
- So the exact pH is right there.
- So our pH of hydrofluoric acid is equal to 1.49.
- So once again, pretty low pH.
- But notice, this pH-- this is 1.5 molar of
- hydrofluoric acid.
- While 1 molar of hydrochloric acid was even more acidic.
- It had a pH of 0.
- So even though we put more acid here, we didn't get our
- pH as low as when we used only 1 molar of hydrochloric acid.
- Anyway, see you in the next video.
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