Ka and Kb of compounds - Online Tutor, Practice Problems & Exam Prep

Ka and Kb

We're going to say here that associated with any weak acid or any weak base, we have our \(Ka\) and our \(Kb\) values. So these are equilibrium constants for weak acids and weak bases respectively. We're going to say \(Ka\) represents the acid dissociation constant for a weak acid and it measures the strength of weak acids. And we're going to say here that the \(Kb\) is representative of our base dissociation constant and it measures the strength of weak bases. We're going to say in general, the higher your \(K\) value is, then the lower your \(pKa\) value is. This translates to a stronger acid. That means that you'll produce more \(H^+\) concentration. It's going to increase the concentration of \(H^+\). We're going to say here that generally speaking, weak acids have \(Ka\) values less than 1. Weak bases use \(Kb\) instead. Their \(Kb\) values are less than 1. Now, because their \(Ka\) and \(Kb\) values are less than 1, that means we do not make an appreciable amount of products. So very little product would be made when our weak acid dissolves in solution or when our weak base dissolves in solution.

Now if we're talking about weak acids and weak bases, we're going to say that they establish equilibriums because they're weak. So for a weak acid, here it's the acid. Water here will act as the base. Based on the Bronston Lohrey theory, weak acids donate \(H^+\) to the base creating \(A^-\) and \(H_3O^+\) as our products. Here, we establish an equilibrium which is the reason for the double arrows. Now, weak acids use \(Ka\). \(Ka\) is an equilibrium constant and like all other equilibrium constants, it equals products over reactants. Remember, we ignore solids and liquids. Based on our equilibrium expression, it would be \(\frac{A^- \times H_3O^+}{\text{Reactants}}\). Again, we ignore the water here because it is a liquid.

For weak bases, here it's the base. Water now acts as the acid. It's going to donate an \(H^+\) over to the base because according to Bronston Lohrey, the acid donates an \(H^+\) over to the base, therefore creating \(\text{OH}^-\) as my products. Weak bases use \(Kb\) as their equilibrium constant. Also, it equals products over reactants. Again, we ignore water because it is a liquid. So these would be our equilibrium expression for a weak acid and for a weak base.

Later on, we'll learn that through these equilibrium expressions, we're able to set up an ice chart to determine the pH or \(pOH\) of a weak acid and a weak base respectively. Now, \(Ka\) and \(Kb\) are connected to each other by the formula \(Kw = Ka \times Kb\). We've seen \(Kw\) in the past before. Remember, this represents our ion product constant for water and remember at \(25^\circ\)C it equals \(1.0 \times 10^{-14}\). This means that at \(25^\circ\)C, if I know \(Ka\), I can determine \(Kb\) because we'll know what \(Kw\) is and vice versa. If I know \(Ka\), I can find \(Kb\) because again we know what \(Kw\) is. Keep in mind this equation here to help you convert from either \(Ka\) to \(Kb\) or vice versa.

Now that we've looked at our descriptions of \(Ka\) and \(Kb\), try to attempt this example question here on the bottom of the page. If you get stuck, don't worry. Just click on to the next video and see how I approach that same question, which relates acid strength between these two acids given in an example on the bottom of the page.

Ka and Kb

Here it says, "Consider 2 aqueous solutions of equal concentration. Which statement is true?" We have chlorous acid, which has a Ka value of Ka1.1×10-2 and phenol, which has a Ka value of Ka1.3×10-10. Alright, so both of these acids here would react with water. They both would donate an H⁺ to water. So, we'd have ClO₂^- and H₃O⁺ being created here. And then here, this phenol would also react with water, donates an H⁺ to water to produce C₆H₅O^- plus H₃O⁺.

Now, remember, the higher your Ka value, the stronger the acid. What does that mean? Well, the higher the Ka, the stronger the acid, the more it dissociates into these ions as products. So basically, the higher your Ka, the more of these you'll produce, the more products you'll produce. And since chlorous acid has a higher Ka value, that means you're going to produce a higher amount of these products. All right. So now, looking at that, let's look and see which statement is true.

• Chlorous acid produces more H₃O⁺ than phenol. That is true because it has a higher Ka value.
• Chloric acid is basic compared to phenol? No, it's more acidic because its Ka is larger.
• Chlorous acid produces less H₃O⁺ than phenol? No, again it produces more H₃O⁺ because its Ka is larger, it's more acidic.
• Chlorous acid is a strong acid. No, we said earlier that weak acids have Ka values less than 1. So, this would be a weak acid.
• Chloride ion produces more OH^- than our phenoxide ion. All right. So, these are the conjugate base forms of those acids. So let's think about this. So, you have ClO₂^- is a stronger acid than phenol. If you're stronger as an acid, that means you're going to be weaker as a base. So, ClO₂^- is a weaker base than C₆H₅O^-. Because remember, if you're strong as an acid, you're weaker as a base. If you're weaker as a base, that means you produce less OH^-. So, chloride wouldn't produce more OH^-, it would produce less because it is the weaker base of the two.

So, out of all the choices, only option A would be true. So, just remember, the higher the Ka, the stronger the acid will be, the more it will produce the desired ions as products. But the stronger you are as an acid, the weaker you will be as a base. Remember, you can't be incredibly strong both ways. Pick a side; you're weaker on the opposite. So, the stronger the acid, the weaker the base from would be.

Ka and Kb Calculations 1

So here it asks, which of the following compounds has the strongest conjugate acid? Now realize that there is give and take. If you exist as one of the strongest acids, that means you're going to exist as also one of the weakest bases. So strongest conjugate acid here is really saying weakest base. And we're looking at the weakest base because all of these compounds, which is the reason they have K b values associated with them. Remember, K b values are associated with weak bases. Since we're looking for the strongest conjugate acid, that really means that we are looking at the weakest base. The weakest base would have the smallest K b value. If we look at all the choices, the one with the smallest K b value would be the last option here. Option D would be our correct choice. Remember, if you're incredibly strong one way, you're incredibly weak the opposite. The strongest acid would give us the weakest conjugate base. Then we'd say here, the strongest base would give me the weakest conjugate acid. Just remember, the opposite duality that exists between acids and bases in order to answer a question like this.

Ka and Kb Calculations 1

At 0 degrees Celsius, the ion product constant of water is 1.2×10-15. Now, the pH of pure water at this temperature is as follows:

So, we're discussing the ion product constant of water, or \( K_W \). Recall that \( K_W = [\text{H}_3\text{O}^+] \times [\text{OH}^-] \). They indicate that \( [\text{H}_3\text{O}^+] = [\text{OH}^-] \) in pure water, implying both concentrations are equal, hence denoted as \( x \). Thus, the equation simplifies to \( x^2 \). At this temperature, \( K_W = 1.2 \times 10^{-15} \). While typically \( K_W = 1.0 \times 10^{-14} \) at 25 degrees Celsius, it changes with temperature alterations.

We set \( x^2 = 1.2 \times 10^{-15} \). Solving for \( x \) (the concentration of \( \text{H}_3\text{O}^+ \)), by taking the square root, we find \( x = 3.464 \times 10^{-8} \). Now, to find the pH:
\( \text{pH} = -\log[\text{H}_3\text{O}^+] \).

Plugging in the calculated \( x \), the resultant pH is \( 7.46 \). This final pH value is crucial, as it helps confirm that we correctly addressed the problem without worrying about significant figures here specifically, but aiming to correctly apply the given constants and calculations.

Recall, if the temperature deviates from 25 degrees Celsius, the value of \( K_W \) shifts accordingly, which must be taken into account. Such values are given as needed; thus, memorization is not mandatory, but understanding how to apply them is crucial.

(CH₃)₂NH is a weak base. Which equilibrium corresponds to the acid dissociation constant K_a for (CH₃)₂NH₂⁺? 

• A. (CH₃)₂NH (aq) + H₂O (l) ⇌ (CH₃)₂N⁺ (aq) + H₃O⁺ (aq)
• B. (CH₃)₂NH₂⁺ (aq) + H₃O⁺ (aq) ⇌ (CH₃)₂NH (aq) + H₂O (aq)
• C. (CH₃)₂NH₂⁺ (aq) + H₂O (l) ⇌ (CH₃)₂NH (aq) + H₃O⁺ (aq)
• D. (CH₃)₂NH₂⁺ (aq) + OH⁻ (aq) ⇌ (CH₃)₂NH (aq) + H₂O (aq)
• E. (CH₃)₂NH₂⁺ (aq) ⇌ (CH₃)₂NH (aq) + H⁺ (aq)