0:28

When we look at water what we see is that water can actually undergo

Â auto-ionization. Two water molecules

Â can react and produce an H_30+ ion and hydroxide ion.

Â Notice that this process is at equilibrium and the equilibrium lies far

Â to the left.

Â We only get a very small amount to be H_3O+ ion and and the hydroxide ion.

Â We can also show this in a little bit about shorthand notation by showing a

Â single water molecule

Â going to H+ a quiz and OH- aqueous.

Â Remember that we cannot actually have H+ ions free in solution.

Â If we write a H+, what we really mean is that we have

Â H_3O+ present. If we look at some of the properties of water

Â one thing we notice is that it is amphoteric. This means that it can act as

Â an acid or a base.

Â Just as when we talked about the Bronsted-Lowry definition of acids and bases

Â we said whether water acts as an acid or base depends on the identity and the

Â other reactant.

Â K_w is in equilibrium constant specifically we call it the ionic product constant

Â for water.

Â We can also call it a dissociation constant, but regardless of what we call

Â it, it is still an equilibrium constant.

Â The W in the subscript let us know that this is for water.

Â We can still write ab equilibrium expression for this just as we would

Â for any other reaction.

Â So how do we write the equilibrium expression?

Â Well we have a reaction in here we are using our shorthand notation but either

Â one will show us the same thing.

Â We have H_2O liquid in equilibrium with H+ an OH- ions.

Â When I'm writing an equilibrium constant expression or a law of mass action,

Â I am writing the concentration that the products raised to their powers

Â which we have coefficients have one for both H+ and OH-.

Â So we don't have any powers in the expression. On the bottom in the

Â denominator we put the concentration of our reactants.

Â However, we remember that we exclude both are solids and are pure liquids.

Â So we do not include water in our denominator.

Â So the equilibrium expression or the ion product expression in this case

Â is K_w equals H+ times

Â OH-. Another thing that's important to remember is that the value of K_w.

Â Notice that this is only true at 25 degrees Celsius that K_2 equals

Â 1.00 time 10 ^ -14.

Â As we change temperatures we change the value of k

Â that equilibrium constant just as the values of equilibrium constants for

Â other reactions also change when we have a change

Â in temperature.

Â Let's look and see what we can find about the values at the H+

Â and OH- concentration for a sample pure water

Â given that we know the value of K_w. So we are at 25 degrees Celsius

Â are K_w value is 1.0 x 10 ^-14.

Â That equals the H+ concentration times the OH- concentration.

Â If we are looking at a sample a pure water are H+ concentration must be

Â equal to our OH- concentration because the balanced reaction

Â and so we represent them both as X therefore we end up with 1.0 x 10 ^-14

Â equals X^2. When I solve for X, I get 1.0 x 10 ^-7.

Â That is going to be equal to the concentration of H+

Â as well as the concentration ever OH-

Â in a sample of pure water. We are also going to look at later about what happens

Â when we are not at 25 degrees Celsius when we have a different value

Â K_w and what that means for the concentrations of H+ and OH-

Â both in pure water

Â as well as in either an acidic or basic solution.

Â First, let's look at example of how we find OH-

Â if we know the H+ concentration. So for example if we have a solution that has an

Â H+ concentration equal to 3.5 x 10^ -5

Â we know we are at 25 degrees so we know that out K_w value is going to be

Â equal to 1.0 x 10 ^-14.

Â We can set up our expression 1.0 x 10 ^-14

Â equals 3.5 x 10 ^ -5

Â times are hydroxide concentration thats are unknown that we're trying to find.

Â When I divide both sides by 3.5 x 10 ^ -5

Â I end up with 2.9 x 10 ^ -10 equals the OH- concentration.

Â I could do a similar calculation if I knew the OH- concentration

Â and needed to find the H+ concentration, but at 25 degrees Celsius

Â this expression will always be true.

Â 1.00 x 10 ^-14 equals

Â H+ times the OH- concentration.

Â 5:37

Most the time are when we looking at solutions we're looking at them at 25 degrees Celsius

Â and so we most commonly use this K_w value 1.0 x 10 ^-14

Â However, if we decrease the temperature

Â we see that the value of K_w changes or if we increase the temperature

Â we see that the value K_w changes as well.

Â You should know the value at 25 degrees Celsius but if you doing calculations at

Â a different temperature

Â then it's okay to be able to look at these values. Now note that some of these values

Â are not written in the correct format

Â for scientific notation because we wanted to be able to compare the numbers

Â and have the powers be all the same.

Â 6:27

So when we are at 25 degrees Celsius if we have a sample a pure water

Â we know that it has an H+ concentration at 10 x 10 ^ -7.

Â And what we want to be able to do is find the pH from that

Â and so we see the pH is equal to the negative log at the H+ concentration.

Â So we can find that the pH of water

Â when the concentration of H+ is 1 x 10 ^ -7.

Â The pH is equal to 7.

Â Now when the pH is equal to 7

Â we know that the pOH is also equal to 7 and we can do this two different ways.

Â We could take the pOH equals minus log of OH-.

Â Remember if the H+ concentration is equal to 1 x 10 ^ -7

Â that the OH- concentration is also equal to the 1 x 10 ^ -7.

Â So I can take the negative log of 1 x 10 ^ -7 and

Â find the pOH value

Â is seven as well, or we can take advantage of the fact that pH plus pOH

Â equals 14.

Â Now note this only applies at 25 degrees Celsius because our K_w value is equal to

Â 1 x 10 ^ -14.

Â If we had a different K_w value because we read a different temperature

Â then this expression would no longer be true. We would still be able to say pH

Â plus pOH equals

Â plus something, but we have to have a new value instead of 14.

Â Using a pH meter we can easily measure the pH of a solution.

Â However that doesn't directly give us the concentration at the H+ ions.

Â But we can use the pH and be able to calculate that H+ ion concentration.

Â This is our expression for finding the pH from our H+ concentration. This

Â is the relationship that we know:

Â pH equals minus log H+ concentration.

Â Now I'm going to rearrange this to solve

Â for H+ so I have -pH, I divided both sides by -1.

Â So -pH equals the log of H+. Now to get rid of the log I am going to take

Â 10 to both sides.

Â Remember that 10 and log are

Â opposite functions. In the same way that if I divide by six or multiplied by

Â 6 they do opposite things to the number.

Â What I can also see is that X equals

Â 10 to the log of X.

Â The log and the 10 cancel each other out

Â and we are left with X. We can take advantage if this by saying 10^-pH

Â equals 10 to the log of H+ and I can simplify that

Â because those two terms, the effect will cancel each other out

Â so we end up with 10 ^-pH equals the H+ ion concentration.

Â So if I were able to measure the pH of a solution

Â I can then find the H+ concentration.

Â I could also do the same thing with my pOH if I had

Â pOH value, I could find the OH- concentration.

Â Now we want to look at some examples where we are not at 25 degrees Celsius.

Â 9:29

Where we are at a different temperature in this case 40 degree Celsius

Â and out K_w value is 2.92 x 10 ^ -14.

Â Now note that this first part is just about looking at the conversion

Â of H+ from the value of the pH

Â that was measured.

Â You should have found that the H+ concentration was 3.98 x 10 ^-5

Â molar. We can do this by saying 10 ^-pH equals

Â the H+ concentration. This is the expression we derive from the

Â formula for finding the pH from pH equals negative log of

Â H+. Here I put in 10 ^ -4.40

Â equals my H+ concentration

Â and I find that equals to 3.98 x 10 ^-5

Â molar. Now

Â we are going to take this one step further and look at what the OH-

Â concentration of this same solution.

Â But we have to remember that we are not, in fact, at

Â 25 degrees Celsius we have a different value of K_w.

Â So what is the OH- concentration of a solution with a pH of 4.40

Â at 40 degrees noted that the K_w value

Â at that temperature is given.

Â Now we can solve for that OH- concentration. We know that K_w

Â is equal to H+ times the OH- concentration.

Â We know that K_w is equal to 2.92 x 10 ^ -14

Â and we know from our previous problem that our H+ concentration is 3.98 x 10 ^-5

Â molar. So now we can solve for our OH- concentration

Â and what we find is that it's 7.5 x 10 ^ -10 molar.

Â We can go back and check our work by multiplying 3.98 x 10 ^-5

Â times 7.5 x 10 ^ -10 and we should get a value somewhere close to

Â the 2.92 x 10 ^ -14.

Â However we have to ask ourselves why can't we use pH plus pOH equals 14?

Â Remember this is only true at 25

Â degrees celsius. This is only true because K_w at 25 is equal to 1.0 x 10 ^-14.

Â A different value K_w we cannot say pH plus pOH equals

Â 14. Let's look at an example where we look at the OH- concentration given

Â the H+ concentration.

Â We are 25 degrees Celsius so we can remember that K_w equals 1.0 x 10 ^ -14.

Â Once we find those two values we can then find pH

Â and pOH by knowing the H+ concentration pH equal -log H+.

Â We could find the pH value for pOH we can do the same thing

Â 9.54 and what we do notice is that 4.46

Â plus 9.54

Â is equal to 14. This works out because we're at 25 degrees Celsius.

Â When we're looking at calculations involving logs we have to be careful

Â about our significant figures.

Â Note that this value the 3.5

Â has two significant figures that will be the number of digits in our mantissa

Â or the number of decimal places in our answer.

Â So we have two sig figs here.

Â We have two decimal places here, but as a result we end up with 3 sig figs in our

Â answer. So the number of sig figs equals the number of decimal places.

Â In the next module we will talk about acid strength and compare strong and weak acids.

Â