Recall that strong acids and strong bases are considered to be strong electrolytes. That means that when we place them in solution, they will completely ionize within the solvent. Now, we're going to say in general, remember we also said that the larger the Ka value for an acid then the stronger the acid will be and the greater the concentration of H^{+} ions. We could also say the larger the Kb value, the stronger the base and therefore the larger the OH^{-} concentration. Now here, for example, we have hydrochloric acid as an example. Here when it's thrown into water, our solvent, it completely ionizes into H^{+} ion and Cl^{-} ion. Because it is a strong acid, there's a 100% ionization, so the initial concentration of my acid represents the final concentrations of the ions formed because again, it's a 100% ionization. Now, strong acids and strong bases are both strong species, of course. If we're taking a look at acids, the fact that these acids are all strong, they're going to have Ka values that are greater than 1. We can easily see that when we look at each of them, all of them give me a value greater than 1 or equal to 1 in the case of chloric acid. Now, we would normally say that if we take a look at the concentration of my strong acid or strong base, we can just take the negative log of that concentration to find pH or pOH respectively. Now that's not entirely true. Even with these strong species, we have to be mindful of the initial concentration given to us. Based on the initial concentration, have to take different approaches and understand different methods to find the correct pH and pOH. So take a look at the next video and see how we break down the different concentrations of strong acids and strong bases to help us determine the best course of action in order to find our pH or our pOH respectively.

- 1. Chemical Measurements1h 50m
- 2. Tools of the Trade1h 17m
- 3. Experimental Error1h 52m
- 4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
- 6. Chemical Equilibrium3h 41m
- 7. Activity and the Systematic Treatment of Equilibrium1h 0m
- 8. Monoprotic Acid-Base Equilibria1h 53m
- 9. Polyprotic Acid-Base Equilibria2h 17m
- 10. Acid-Base Titrations2h 37m
- 11. EDTA Titrations1h 34m
- 12. Advanced Topics in Equilibrium1h 16m
- 13. Fundamentals of Electrochemistry2h 19m
- 14. Electrodes and Potentiometry41m
- 15. Redox Titrations1h 14m
- 16. Electroanalytical Techniques57m
- 17. Fundamentals of Spectrophotometry50m

# General Approach to Acid-Base Systems - Online Tutor, Practice Problems & Exam Prep

Strong acids and bases are strong electrolytes that completely ionize in solution, leading to significant concentrations of H^{+} or OH^{-} ions. For concentrations ≥ 10^{-6} M, pH or pOH can be calculated using the negative logarithm of the concentration. However, for concentrations < 10^{-8} M, the pH is neutral at 7. Between 10^{-6} and 10^{-8} M, the autoionization of water must be considered to accurately determine pH and pOH. Understanding these principles is crucial for acid-base reactions and calculations.

## The Systematic Approach

### Systematic Approach

#### Video transcript

### Systematic Approach

#### Video transcript

We've been under the assumption that whenever we're given the concentration of a strong acid or strong base, we could take the negative logs of those concentrations to find pH or pOH respectively. Now, in actuality, when calculating the pH of a solution, we must take into consideration the concentration of the strong acid or the strong base because, depending on the concentrations given, we'll have to use different methods to find the correct pH and pOH respectively. Now, here we split it up into 3 categories.

Here, for the first one, when the concentration of your strong acid or strong base is equal to or greater than 10 -6 molar, that means that there's a significant amount of those two species that we can just simply look at their concentrations and take the negative log. For example, here we have nitric acid. It completely ionizes into H+ ion and nitrate ion. Because the concentration of 1.5×10-3 molar is greater than 10-6 molar, we can just simply take the negative log of this concentration because, remember, nitric acid, being a strong acid, completely breaks up into H+ ion. This would be a 100% generation of this ion. I would take the negative log of that concentration, and that would give me my pH, which comes out to 2.82.

Here, we have sodium hydroxide. Again, the concentration, as long as it's equal to or greater than 10 -6 molar, we have complete ionization. Here, we have a 100% generation of this OH- ion. Here, we can just simply take the negative log of this concentration, but now it gives me pOH. So I plug that value in. That'd give me 3.12. And then remember, at this point, pH equals 14 minus that value. We'd get 10.88 at this point. As long as my concentration is equal to or greater than 10 -6 molar, the concentration of the strong acid and strong base is significant enough to change the pH of my solution. I can simply look at their concentrations and take the negative log to find pH or pOH respectively.

Now, when the concentration they have is equal to or less than 10 -8 molar, and really we're going to say when it's less than 10 -8 molar, then we're going to say that the concentrations are not significant enough. Therefore, the pH will be 7. That's because there isn't enough strong acid or enough strong base to change the pH of the solution as a whole. Here, we have 10 -11 and 10 -9, they aren't significant enough so both solutions would be neutral. Again, it's when concentrations are less than 10 -8 molar.

Finally, when you have your concentration between 10 -6 molar to 10 -8 molar, we're going to take a systematic approach to finding our pH. That's because, at these concentrations, they're not large enough but not small enough, so they will have some effect on changing the pH of our solution. Here, we have to be mindful that between these concentration values, we have to take into account the auto-ionization of water. Remember, the auto-ionization of water would produce some H+ ion and some OH- ion. These concentrations would actually contribute to the overall concentrations of H+ and OH-. Taking them into consideration is key to getting the correct pH for our strong acid and the correct pOH for our strong base. Remember, when it comes to strong acids and strong bases, they are strong electrolytes that completely ionize in water, but we have to be careful when looking at their concentrations. Depending on where their concentration lies means we'll have to take one of three possible routes to find pH or pOH respectively. Just keep in mind some of the tips that we went over here. As we continue with this topic, keep in mind the strategies needed to find your pH or pOH.

Whenever the concentration of your strong acid or strong base is between 10^{-6} to 10^{-8} M a systematic approach may have to be used to determine pH or pOH.

### Systematic Approach Calculations

#### Video transcript

Here we need to determine the pH of a \(3.5 \times 10^{-8}\) molar HBr (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 \(3.5 \times 10^{-8}\) 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: 1 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—Remember, 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: \(3.5 \times 10^{-8}\) molar.

Now, this is important because this will eventually lead us to the correct concentration for our \(H^+\) ion. So now we're going to say here \(Br^- = 3.5 \times 10^{-8}\) molar. So we can say now that \(H^+ = 3.5 \times 10^{-8}\) molar because it equals bromide ion plus the concentration of \(OH^-\) which we don't know so that's equal to \(x\). Now we can say from this information that \(K_w\), which is our dissociation constant for water, equals \(H^+ \times OH^-\). Plugging the values that we know, \(K_w = 1.0 \times 10^{-14} = 3.5 \times 10^{-8} + x\). We don't know what \(OH^-\) is, so it's also \(x\).

From this set of values, we can figure out what \(x\) will be. We are going to distribute this \(x\), that's going to give me \(1.0 \times 10^{-14} = 3.5 \times 10^{-8} \times x + x^2\). That \(x^2\) has the largest power for the \(x\) variable. Rewriting our equation now gives us \(x^2 + 3.5 \times 10^{-8} \times x - 1.0 \times 10^{-14}\). That is our equation and from it, we'd say that this would be \(a\), \(b\), and \(c\). We're going to use the quadratic formula: \(-b \pm \sqrt{b^2 - 4ac} \over 2a\). So let's plug in the values that we know. So that'd be \(-3.5 \times 10^{-8} \pm \sqrt{(3.5 \times 10^{-8})^2 - 4 \times 1 \times -1.0 \times 10^{-14}} \over 2 \times 1\). Now, when I solve for all of this here, I'll get \(-3.5 \times 10^{-8} \pm 2.03039 \times 10^{-7} \over 2\).

And realize here that our result will give 2 x variables, one that's positive and one that's negative: \(2.03039 \times 10^{-7}\). So my 2 possible answers for \(x\), you'll get \(8.4 \times 10^{-8}\) molar or you'll get \(x = -1.19 \times 10^{-7}\) 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 just simply \(x\). You could take that value and plug it in, and it represents \(OH^-\) concentration, which means that if I take the negative log of it, I'll get \(pOH\). So when I take the negative log of it, I get \(pOH = 7.08\). And here, if we know the \(pOH\), we know 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 \(6.92\) is pretty high of a pH, but that's because the concentration of HBr is very diluted.

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 see. 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 do the systematic approach to calculating the correct concentration of \(H^+\) and from there determining 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 in similar ways. Remember what is the complete ionization of the strong base going to result? 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? 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 guys were able to follow along with this example on calculating the pH of a strong acid using the systematic approach.

Determine the pH of a 6.7 x 10 ^{-8} M NaOH.

### Here’s what students ask on this topic:

What is the general approach to calculating the pH of strong acids and bases?

To calculate the pH of strong acids and bases, you need to consider their complete ionization in solution. For concentrations ≥ 10^{-6} M, you can use the negative logarithm of the concentration to find pH or pOH. For example, for a strong acid like HCl with a concentration of 1.0 × 10^{-3} M, the pH is calculated as:

$\mathrm{pH}=-\mathrm{log}(1.0\times {10}^{-})=3$

For concentrations < 10^{-8} M, the pH is neutral at 7. For concentrations between 10^{-6} and 10^{-8} M, you must consider the autoionization of water, which contributes to the overall H^{+} and OH^{-} concentrations.

How do you calculate the pH of a strong acid with a concentration less than 10^{-8} M?

For strong acids with concentrations less than 10^{-8} M, the concentration is not significant enough to affect the pH of the solution. In such cases, the pH of the solution is neutral, which is 7. This is because the autoionization of water (H_{2}O ⇌ H^{+} + OH^{-}) dominates, and the concentrations of H^{+} and OH^{-} ions from water are both 1.0 × 10^{-7} M, leading to a neutral pH.

Why is the autoionization of water important when calculating pH for certain concentrations of strong acids and bases?

The autoionization of water is important when calculating pH for strong acids and bases with concentrations between 10^{-6} and 10^{-8} M. In this range, the concentrations of H^{+} and OH^{-} ions from the acid or base are not large enough to ignore the contribution from water's autoionization. The autoionization of water produces H^{+} and OH^{-} ions at 1.0 × 10^{-7} M each, which must be added to the ion concentrations from the acid or base to accurately determine the pH or pOH.

How do you calculate the pOH of a strong base like NaOH?

To calculate the pOH of a strong base like NaOH, you need to consider its complete ionization in solution. For concentrations ≥ 10^{-6} M, you can use the negative logarithm of the concentration to find pOH. For example, for NaOH with a concentration of 1.0 × 10^{-3} M, the pOH is calculated as:

$\mathrm{pOH}=-\mathrm{log}(1.0\times {10}^{-})=3$

To find the pH, use the relationship:

$\mathrm{pH}=14-\mathrm{pOH}=14-3=11$

What is the significance of the Ka and Kb values in determining the strength of acids and bases?

The Ka (acid dissociation constant) and Kb (base dissociation constant) values are crucial in determining the strength of acids and bases. A larger Ka value indicates a stronger acid, as it signifies a greater tendency to donate H^{+} ions. Similarly, a larger Kb value indicates a stronger base, as it signifies a greater tendency to donate OH^{-} ions. For strong acids and bases, Ka and Kb values are typically greater than 1, indicating nearly complete ionization in solution. These values help predict the behavior of acids and bases in various chemical reactions and are essential for accurate pH and pOH calculations.