In this module we will look at how we found the pH of both strong and weak acid solutions. Our objective in this unit is to calculate the pH are both strong and weak acid solutions because we have different procedures for each. Finding the H+ concentration of a strong acid is fairly easy. So finding the pH of that same solution is a straightforward process. When we look at a strong acid we know that it is a strong electrolyte meaning it ionizes is completely. HCl goes to H+ and Cl. If we have a solution that is 0.050 molar HCl, as in that is the concentration that we measured then we can know that the complete dissociation, or complete ionization will result in all other HCl being converted into H+ and Cl- ions So if we have a complete dissociation and we know the initial concentration of HCl. Then we can also find the final concentration of H+ because it will be the same as the initial. Therefore H+ is going to be equal to 0.050 molar. Once we know the H+ concentration finding the pH is easy. pH equals negative log a the H+ concentration. So in this case pH is equal to negative log of 0.050 molar. And out pH will be equal to 0.30. Note that I have too sig figs in my original concentration. So I have two decimal places in my answer. Finding the H+ concentration of a weak acid is a little bit different. Because we don't get a complete dissociation or ionization. We have to figure out what fraction for acid dissociate into the ions. If we look at our reaction we have HA yields H+ plus A- and this is an equilibrium process. If you're not sure if something is a weak or strong acid the presence of that equilibrium arrow indicates that it is in fact a week acid. If we look at our initial concentration of HA we see that is not equal to the final H+ concentration as it was for the
indicates that it is in fact a week acid.
If we look at our initial concentration of HA we see that is not equal to the final H+ concentration as it was for the strong acid. Because not all of our HA is divided or dissociated into the H+ an A-. We have to look a K_a the acid dissociation our ionization constant for the acid. And remember this is just an equilibrium constant for the acid. So our procedure is going to be much more like what we did in the previous unit talking about equilibrium. We are going to use an ICE table to solve for that unknown concentration. Let's look at an example. What is the pH about 0.200 molar solution of HF. And were given that the K_a value is equal to 3.5 x 10^-4. Here we have HF, aqueous phase, plus F-. I set up my ICE table remember, Initial, Change, and Equilibrium. I know my initial concentration of HF is 0.200 molar because that's what was given in the problem. I know my initial concentrations of H+ an F- are zero. Because we have not had any dissociation to occur initially. Then I look at my change row I'm going to lose some amount of HF and I'm going to gain the same amount of H+ plus an F- because if the coefficient it's all a one to one to one ratio. Then I see that for my equilibrium row I have 0. 200 - X and X. Now, I can set up my law of mass action plug in my terms from the equilibrium row and solve for my concentration. So let's look at that how we would do that. So here we have H+ and F- are both equal to x. So I can say that equal to x^2 and the HF is going to be 0.200 - x.
So here we have H+ and F- are both
equal to x. So I can say that equal to x^2 and the HF is going to be 0.200 - x. On the bottom equal to 23.5 x 10 ^-4. Just as we did in the earlier equilibrium problems. We want to make the simplifying assumption that X is much much less then 0.200. Given the small value K_a this is probably a reasonable assumption, but we would still need to do the calculation just to make sure. So now we got out X's is less than 0.20. We simplified our expression so let's rewrite what we have left. Now I have lost the of the -X because we have simplified. Now I can multiply both sides by 0.200. And when I do that I end up with 7.0 times 10 ^ -5 times X^2. I can take the square root of that number and found that X equals 8.4 times 10 ^-3. That will be equal to my H+ concentration. From that, if we wanted, that we could find the pH value are we go on to find the OH- concentration using what we know about K_w. finding the pOH- value. Another way of looking at the acid strength is to look at the percent ionization of weak acids. What we can look at is the H_3O+ concentration at equilibrium or remembering that the same thing we've been calling the H+ concentration. those mean exactly the same thing. Divided by the initial HA concentration and multiplied by 100. The bigger the value of the percent ionization the stronger the acid and vice versa. The smaller the percent ionization the weaker the acid. Let's look at an example of how we can use this information to figure out the pH of the solution. Here we have HA in equilibrium with H+ and A-. It is 0.350 molar weak acid that is 13.2 percent ionized and we want to find the pH of the solution. We have percent ionization is equal to H+ at equilibrium over HA initial times a hundred. I could plug in the values I know is equal to H+ at equilibrium over HA
initial times a hundred. I could plug in the values I know 13.2 percent equals H+ it equilibrium over 0.350 initially times 100. Now I can rearrange and solve for my H+ concentration at equilibrium and find that is 0.462. That gives me the concentration it does not yet give me the pH value. I have to take the negative log of that number to find the pH value which is 1.335. We have three-digit in our decimal place and the mantissa of the log because we have three significant figures in our concentration. When I look at a mixture of strong and weak acid I have a couple of things to consider. First of all the strong acid
When I look at a mixture of strong and
weak acid I have a couple of things to consider. First of all the strong acid completely dissociate or ionize. Because I go from HCl to producing H+ and Cl- and all of my HCl will dissociate. So I have a fairly large concentration of H+ present. When I look at the HA, if I just had the weak acid by itself I would see the formation of some H+ and some A-. This number even by itself is going to be a small number. So compared to this H+ that I get from the acid. However because the week acid process is in equilibrium process and we've got our equilibrium arrow there. It is an equilibrium process. What we're going to see is that the present at the H+ from the strong acid ionization will actually drive the equilibrium towards the left for the weak acid and as a result the small amount of H+ I actually had present from dissociation the HA will actually get smaller in this mixture with a strong acid. So what I find is that the primary H+ source will be the strong acid and the amount of H+ produced from that weak acid is so small that it becomes insignificant and I can ignore it. Now from looking at a mixture of two or more weak acids. I have multiple reactions going on and just like I did with the strong and weak acids I'm gonna find the strongest have those acids. if there are guys are fairly different from one another say we have K_a values of 10^-3 10 ^ -5 and 10 ^-9 those K_a values are far enough separated from one another that I really only have to look at the one that has the K_a of 10^-3. And I will do all my calculations as if that were the only one at present. Now I will a look at an example which one of the following is associated with the acid? Now note these all have K_a values so they are all weak acids. So we're asking which one is the strongest week acid at these four. Hopefully you answered the first 11.9 x 10 ^-4. This is the largest the K values for these four. Therefore, it will be the strongest acid of these. We talked a lot about acids both strong and weak. In the next module what we are going to look at is bases.
