K A 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 K A 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 K A 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 plus by donating H plus, it becomes F minus and water gaining the H plus acting as the base becomes H +30 plus. Here K A 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 hydro ions divided by the concentration of Hydrofluoric acid. This expression will be equal to the value of K A, the K A of our weak acid. This value happens to be 6.3 times 10 to the negative four. Now, what can we talk about in terms of acid based strengths in relation to K A? Well, we can say here the stronger the acid, the higher the K A value will be. We can say here that weak acids tend to have K A values less than one and strong acids tend to have K A values greater than one. Now that we've talked about the K A value. Let's look at KBKB 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 plus away. Ammonia accepts an H plus and becomes the ammonium ion because water loss in H plus, it becomes the hydroxide ion. KB just like K A 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 hydro 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 five. Here, the stronger the base is then the higher its KB value will be, we say here that weak bases tend to have KB values less than one and strong bases tend to have them greater than one. Now, here it's strong acids and bases have an association constant associated with them as well. It's just that for strong acids, their K A values are greater than one. So, so greater than one that we don't need to talk about them. Strong bases have KB values much greater than one. So again, we don't talk about their KB values either K A, we mainly stick with weak acids. KB, we stick with weak bases, right? So just remember this, these are just association 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 K A values here, assumed temp is 25 °C. 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 °C because this is seen as standard temperature or room temperature where we have a good fix on what the K A value for a weak acid would be. If we take a look here, we have four different weak acids with K A values. We know all of them are weak because they have K A values less than one. We want the strongest acid and remember the higher your K A value, the stronger the acid will be. And we take a look here. This is to the negative 10, negative 14, negative four and negative four. So these two are out. So it's either C or D here, this is 4.6 times 10 to the negative four. This one's only 1.4 times 10 to the negative four options. See is the correct answer. Nitrous acid has a K value of 4.6 times 10 to the negative four 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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K A 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 KB. When we take the negative log of KWK A and KB, then it becomes 14 equals PK A plus PKB. Now what exactly do I mean by conjugate pairs? Well, here we have nitrous acid which has a K value of 4.6 times 10 to the negative four. Its conjugate base is the night white ale. We can use this formula here to convert K A 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 K A and KB and their product equals KW. Now here, when we talk further about uh PK A and PKB, we can say for acids we have PK A remember P just stands for negative log. So PK A means negative log of K A. If we know the PK A, we can find K A as well because here we can say that K A equals 10 to the negative P K A. 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 K A and the higher the K A, the lower the PK A, a strong acid tends to have a K A greater than one and a PK, a, less than one. A weak acid tends to have a K A less than one and a PK A value greater than zero for basics. 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 P A B in terms of strength, we can say the stronger the base and the higher the KB and the lower the PK A strong bases tend to have K BS greater than one and PK A is less than one. Weak assets tend to have KB values less than one and PKB values greater than zero, right? So just remember, we have these connections between our acid association, constant K A and the base associated 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 KAKBPK A 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 four. And it's a medication used to reduce pain, fever and inflammation. Now, here it says to calculate the KB 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 acid, I'm denoting its acid form and this ends with eight which represents its conjugate base form. So we have a weak acid which has its K value and we're looking for the KB of its conjugate base form. In this case, we'd say KW equals K A times KB. Here, we're assuming the reaction is happening at 25 °C since temperature is not given to us. So KW is 1.0 times 10 to the negative 14 K A is 3.3 times 10 to the negative four. We're looking for KB. So divide both sides here by 3.3 times 10 to the negative four. So those cancel out when you punch it into your calculator, you'll get KB equals 3.0 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: