Here we need to determine the pH of a \(3.5 \times 10^{-8}\) molar HBr or hydrobromic acid solution. Now remember, here the concentration falls between \(10^{-6}\) to \(10^{-8}\). So we have to make sure, if necessary, to use the systematic approach. When we take the negative log of this strong acid, we'll take the negative log of \(3.5 \times 10^{-8}\) molar. That's going to give me a pH of 7.46.
We can clearly see that this is a basic pH which makes no sense because we have a strong acid. Because we have a strong acid, we should get a pH that's more acidic. This means that we definitely need to do a systematic approach to calculating our correct concentration of \(H^+\) ion and from there determining our pH. Now we know that HBr is a strong electrolyte so it's going to break up completely into \(H^+\) ions and \(Br^-\) ions. Remember, within this region of \(10^{-6}\) to \(10^{-8}\), we have to compete with the autoionization of water.
Remember, in the autoionization of water, we have 2 water molecules. One will act as an acid to produce \(H^+\) and the other one will become \(OH^-\). Here, these are all the ions that are present within our solution. From these ions, we're going to say that our charge balance will take a look at all the ions within the solution. All the positive ions will equal the negative ions.
Here, our positive ion is \(H^+\) ion which comes from 2 sources. It comes from the HBr acid but also comes from the autoionization of water. That's going to equal my negative ions which in this case would be my bromide ion plus my hydroxide ion. That will be my charge balance. Next, we would look at our mass balance and we're focusing on the bromide ion.
We're not taking a look at \(H^+\) ion because it's found in 2 locations, HBr and the autoionization of water. We're going to say here that the mass of bromide ion comes entirely from HBr. Because of that, that means the bromide ion, all of it would have the same concentration as HBr. It'd be \(3.5 \times 10^{-8}\) molar. Now, this is important because this will eventually lead us to the correct concentration for our \(H^+\) ion.
Alright. Now we're going to say here \(Br^- = 3.5 \times 10^{-8}\) molar. \(H^{+} + Br^- + OH^-\) so we can say now that \(H^+ = 3.5 \times 10^{-8}\) molar because it equals the bromide ion plus the concentration of \(OH^-\) which we don't know so that's equal to \(x\). This is what \(H^+\) equals. Now we can say from this information that \(K_w\), which is our dissociation constant for water equals \(H^+ \times OH^-\).
Plugging in the values that we know, \(K_w = 1.0 \times 10^{-14} = (3.5 \times 10^{-8} + x) \times x\). From this set of values, we can figure out what \(x\) will be. What we're going to do now is distribute this \(x\), distribute this \(x\). That's going to give me \(1.0 \times 10^{-14} = 3.5 \times 10^{-8}x + x^2\).
That \(x^2\) has the largest power for the \(x\) variable, so it's our lead term. That means we're going to have to subtract \(1.0 \times 10^{-14}\) from both sides. Rewriting our equation now gives us \(x^2 + 3.5 \times 10^{-8}x - 1.0 \times 10^{-14} = 0\). From it, we'd say that these would be \(a\), \(b\) and \(c\).
We're going to use the quadratic formula, \(-b \pm \sqrt{b^2 - 4ac} / 2a\). So let's plug in the values that we know. So that'd be \(-3.5 \times 10^{-8} \pm \sqrt{(3.5 \times 10^{-8})^2 - 4 \times 1 \times (-1.0 \times 10^{-14})} / 2 \times 1\).
Alright. So now when I solve for all of this here, I'll get \(-3.5 \times 10^{-8} \pm 2.03039 \times 10^{-7} / 2\). And realize here that we'll get 2 \(x\) variables, one that's plus and one that's minus \(2.03039 \times 10^{-7}\). So my two possible answers for \(x\) are \(8.40 \times 10^{-8}\) molar or \(x = -1.19 \times 10^{-7}\) molar. Now remember, concentrations cannot be negative, so this one is automatically dropped out. The concentration of \(x\) is \(8.4 \times 10^{-8}\) molar.
Now, what does this \(x\) represent? If we go back to our expression here, we know that \(OH^-\) is equal to simply \(x\). You could take that value and plug it in, and it represents \(OH^-\). You could also take that \(x\) and just plug it into here and that will give you the concentration of \(H^+\). Here, I'm just going to say that this is equal to \(OH^-\) concentration which means that if I take the negative log of it, I'll get \(pOH\).
When I take the negative log of it, I get 7.008. And here, if we know the \(pOH\), we know the \(pH\) because it's \(14 - pOH\). So that's equal to 6.92. This answer makes more sense. We have a strong acid, and it is a concentration that is not less than \(10^{-8}\).
So it shouldn't be a neutral solution. Definitely should not be a basic solution ever. This pH of 6.92 is pretty high, but that's because the concentration of HBr is very diluted. So remember, this is the approach you need to take, the systematic approach for a strong acid.
Remember the key steps that we did here and apply them to any strong acid that you encounter. Your best course of action should always be to take the negative log of the strong acid to see if the pH makes sense. Here, because it's a strong acid, you should get an acidic pH. If you get a basic pH, that's a strong indication that you need to use the systematic approach to calculate the correct concentration of \(H^+\) and from there determine your pH. Now that you've seen this example, attempt to do the practice question left on the bottom where we have to use the systematic approach for calculating the pH of a strong base.
Again, set it up similarly. Remember, what is the complete ionization of the strong base going to result in? What kind of ions are formed? Also take into consideration the autoionization or self-ionization of water. How do those ions contribute to the pool of ions within my solution? But first of all, just make sure you first take the negative log of the concentration and see if your pH makes sense.
If it doesn't make sense, that is a strong indicator again to use the systematic approach. Hopefully, you were able to follow along with this example on calculating the pH of a strong acid using the systematic approach.