Ka and Kb are equilibrium constants for acids and bases respectively, and they are used to measure the strengths of acids of of weak acids and bases. Now if we take a look here, we can talk about the different types of equilibrium constants in reference to Ka and Kb. We can then show great examples of how they relate to typical types of acid base reactions, and then finally we can talk about the relationships they have between their strong and weak forms. Now, here we're going to say that Ka represents our acid dissociation or ionization constant. A great example here is hydrofluoric acid. It is a weak binary acid. Since it is an acid, it donates H+. By donating h+, it becomes f- and water gaining the h+, acting as the base becomes h three o plus. Here, Ka is an equilibrium constant and like other equilibrium constants, it has a ratio of products over reactants. But remember, we ignore solids and liquids. Here, water is a liquid so it would not be included in our equilibrium expression. Here, k would equal the concentration of fluoride ions times the concentration of hydronium ions divided by the concentration of hydrofluoric acid. This expression will be equal to the value of Ka, the Ka of our weak acid. This value happens to be 6.3 times 10 to the negative 4. Now, what can we talk about in terms of acid base strengths in relation to Ka? Well, we can say here the stronger the acid the higher the Ka value will be. We can say here that weak acids tend to have Ka values less than 1, and strong acids tend to have Ka values greater than 1. Now that we've talked about the Ka value, let's look at Kb. Kb is the base association or ionization constant. It is used for weak bases. Here, a weak base that we see is the Ammonium molecule or ammonia molecule. It is the base, so water acts as the acid to donate an H+ away. Ammonia accepts an H+ and becomes the ammonium ion because water lost in H+ becomes the hydroxide ion. KB just like Ka is an equilibrium constant so it is a ratio of products over reactants. Again, we still ignore solids and liquids. So here would be equal to the hydronium ion concentration times hydroxide ion concentration divided by the ammonia concentration. Here, the Kb value of ammonia is equal to 1.8 times 10 to the negative 5. Here, the stronger the base is then the higher its k b value will be. We say here that weak bases tend to have k b values less than 1 and strong bases tend to have them greater than 1. Now here, strong acids and bases have an association constant associated with them as well. It's just that for strong acids, their Ka values are greater than 1 so greater than 1 that we don't need to talk about them. Strong bases have kb values much greater than 1, so again we don't talk about their kb values either. Ka, we mainly stick with weak acids, kb, we stick with weak bases. Right? So just remember this, these are just dissociation constants related to the weak forms of acids and bases.
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example
Ka and Kb Example
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Identify the strongest acid from the following list of weak acids based on their ka values. Here assume temp is 25 degrees Celsius. Now here we're saying that because K a is just like other equilibrium constants, it's affected by changes in temperature. Here we're saying 25 degrees Celsius because this is seen as standard temperature or room temperature where we have a good fix on what the Ka value for a weak acid would be. If we take a look here, we have 4 different weak acids with ka values. We know all of them are weak because they have ka values less than 1. We want the strongest acid and remember, the higher your ka value, the stronger the acid will be. And we take a look here this is to the negative 10, negative 14, negative 4, and negative 4. So these 2 are out, so it's either c or d. Here this is 4.6 times 10 to the negative 4, this one's only 1.4 times 10 to the negative 4. Option c is the correct answer. Nitrous acid has a k value of 4.6 times 10 to the negative 4 and would represent the strongest weak acid from the options given below.
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Problem
Problem
Hypobromous acid (Ka = 2.8 × 10−9) and hydrocyanic acid (Ka = 4.9 × 10−10) are both weak acids. Determine if reactants or products are favored in the following reaction.
HBrO (aq) + CN− (aq) ⇌ BrO− (aq) + HCN (aq)
a) reactants b) products c) both directions are favored equally d) neither direction is favored
A
reactants
B
products
C
both directions are favored equally
D
neither direction is favored
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Problem
Problem
Identify a Bronsted-Lowry acid with weakest conjugate base.
a) H3BO3 Ka = 5.4 × 10−10
b) HF Ka = 3.5 × 10−4
c) HNO2 Ka = 4.6 × 10−4
d) HClO Ka = 2.9 × 10−8
A
H3BO3 Ka = 5.4 × 10−10
B
HF Ka = 3.5 × 10−4
C
HNO2 Ka = 4.6 × 10−4
D
HClO Ka = 2.9 × 10−8
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concept
Ka and Kb Relationship
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Ka and Kb are related to the following formulas and can only be used for conjugate pairs. Here we're going to say that kw, which remember is our ionization constant for water, equals k a times k b. When we take the negative log of kw, Ka and Kb, then it becomes 14 equals pKa plus pKb. Now what exactly do I mean by conjugate pairs? Well, here we have nitrous acid which has a Ka value of 4.6 times 10 to the negative 4. It's conjugate base is the Nitrite ion. We can use this formula here to convert Ka of the weak acid form to Kb of the weak basic form. So that's what we mean, the weak acid form and its conjugate base form. They're connected to each other by Ka and Kb, and their product equals kw. Now here when we talk further about, pKa and pKb, we can say for acids we have pKa. Remember p just stands for negative log, so pKa means negative log of Ka. If we know the pKa, we can find Ka as well because here we can say that Ka equals 10 to the negative p Ka. These formulas allow us to go between pKa and Ka at whim. Now, here let's talk about strength. Well, here we can say that the stronger the acid the higher the Ka, and the higher the Ka the lower the pKa. A strong acid tends to have a ka greater than 1 and a pka less than 1. A weak acid tends to have a Ka less than 1 and a pKa value greater than 0. For bases, here we're going to say that pKb just means negative log of KB. If we know pKb then we know KB because kb equals 10 to the negative pKb. In terms of strength we can say the stronger the base then the higher the k b, and the lower the p k a. Strong bases tend to have k b's greater than 1 and p k a's less than 1. Weak acids tend to have kv values less than 1 and pkv values greater than 0. Right, so just remember, we have these connections between our acid dissociation constant Ka and the base dissociation constant Kb. We can interchange between them when given the weak acid and its conjugate base. We can also establish relationships in terms of strengths of acids when comparing Ka, Kb, pKa and pKb.
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example
Ka and Kb Example
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Here we're going to say that aspirin also known as Acetylsalicylic acid has a k value of 3.3 times 10 to the negative 4, and it's a medication used to reduce pain, fever, and inflammation. Now here it says to calculate the k b value of its conjugate base form. Now here we know it's a conjugate base form because the names are very similar. The difference is this one ends with ich acid, denoting its acid form and this ends with 8, which represents its conjugate base form. So we have a weak acid which has its ka value, and we're looking for the k b of its conjugate base form. In In this case, we'd say kw equals ka times kb. Here we're assuming the reaction is happening at 25 degrees Celsius since temperature is not given to us, so k w is 1.0 times 10 to the negative 14, k a is 3.3 times 10 to the negative 4. We're looking for Kb. So divide both sides here by 3.3 times 10 to the negative 4. So those cancel out, when you punch it into your calculator you'll get kB equals 3 point 0 times 10 to the negative 11. So this would represent our base association constant for the conjugate base form of aspirin.
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Problem
Problem
Identify which of the compounds is the strongest species.
a) Iodic acid pKa = 0.80 b) Acetic acid pKb = 9.24 c) Formic acid pKa = 3.75 d) Ammonium pKb = 4.75
A
Iodic acid pKa = 0.80
B
Acetic acid pKb = 9.24
C
Formic acid pKa = 3.75
D
Ammonium pKb = 4.75
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Problem
Problem
Determine the pKa given the Kb of the following bases: