Ka and acid strength
How to write an equilibrium expression for an acid-base reaction and how to evaluate the strength of an acid using Ka.
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- At0:26why is the oxygen said to get the plus one charge instead of the hydrogen. If oxygen is more electronegative than the hydrogen, wouldn't it make more sense to say the hydrogen is positively charged. Is this because when you draw these molecules, you usually leave out the hydrogens?(12 votes)
- The oxygen will have a +1 formal charge. To find formal charge, you take the number of valence electrons of a free atom, subtract 1/2 # of shared e-, and subtract #of lone e-. In this case for oxygen in H3O+:
Oxygen has 6 valence e- , has three bonds, and has 2 electrons that fill its octet but aren't involved in bonding. Therefore the formal charge is 6-3-2=+1.(12 votes)
- Are there other noteworthy solvents that don't get included in the Ka equation aside from water?(5 votes)
- Water is usually the only solvent involved in common acid-base chemistry, and is always omitted from the Ka expression. Solvents are always omitted from equilibrium expressions because these expressions relate a constant value (denoted by K followed by a subscript like a or b) to the ratio of the concentrations of products to reactants happening at equilibrium. Solvents cannot be expressed in terms of concentrations as they are the dissolving agent . We only care about the changes happening within the water or any other solvent.(16 votes)
- In the acetic acid and water reaction, can the acetic acid grab a proton from water instead of donating it? Because one of the Oxygen's in the acetic acid has two lone pairs and that would be enough to nab a proton from water, no?
Thanks very much!(6 votes)
- Great question! When we talk about acid and base reactions, reactivity (and acidity and basicity) is all relative. In this particular case, acetic acid usually acts as the acid (the proton donor) because it is much better acid than water. Part of this has to do with the products of this acid-base reaction: the acetate ion, CH3COO-, is pretty good at stabilizing the negative charge using resonance.
You are right that the lone pairs on the oxygens in acetic acid could technically grab a proton from water instead of donating it. If that happened, you'd end up with a +1 formal charge on the oxygen that grabbed the proton, and the water that donated the proton would have become OH-. Since OH- is a strong base (much stronger than acetic acid), it would grab a proton back and hang onto it very tightly. So that is not a super likely scenario due to acetic acid being a much stronger acid than water, or another way to think about it is that water is a better base than acetic acid. I hope this helps!!(16 votes)
- Starting from7:53, the proctor stated that acetic acid is a weak acid due to it's poor H+ donation ability. So that makes acetate a strong base. My question is, why is the acetate ion a strong base when the two oxygen atoms and carbon atom can participate in resonance, or electron de-localization to stabilize the negative charge? Shouldn't resonance make the ion more stable, or am I missing a point here?(6 votes)
- Acetate (CHCOO-) isn't a strong base. The strong bases by definition are those compounds with a kb >> 1 and are LiOH, KOH, NaOH, RbOH and Ca(OH)2, Ba(OH)2, and Sr(OH)2.
Acetate ion is a weak base, but it's a better base than its conjugate acid (acetic acid) is. I think the point is the molecule's ability to either donate OH- or accept H+ because either of these will increase the pH of the solution and is less about resonance.(5 votes)
- Why is cl- a weaker base(2 votes)
- Cl- is a weaker base because Cl is very electronegative and will be unwilling to accept a proton to share its electrons, instead it would prefer to keep the electrons for itself.(5 votes)
- If you were to do the recipricol of the ka (i.e. did concentration of reactants over the concentration of products), would that be your kb? I think that correlates to base strength...(2 votes)
- Ka =(A-)*(H3O+)/(HA)
And Kb = (HB+)*(OH-)/(B)
If it were the same reaction (with an acid and its conjugate base, it would be this way:
And Kb = (HA)*(OH-)/(A-)
Clearly, ka =/ 1/Kb : one has H3O+ and the other has OH- (the others are in the equation, but in very very very small quatities so we ignore them)
For example, HCN is an acid and CN- is its conjugate base. @25°C in aq. state
Ka (HCN) = 6.0 *10^(-10) M and Kb (CN-) = 1.7 *10^(-5) M. So ka =/ 1/Kb
Another intuition is that if an acid reaction is partial, the basic reaction that corresponds should be partial too, but if you inverse Ka you would have a huge Kb, which doesn't make sense.
Plus, since the reaction happens in water, Ka*Kb = Kwater so Ka =Kw/Kb not 1/Kb
Hope this helped ;)(6 votes)
- Around5:30, it was explained that the Ka value was [H3O+] times [Cl-], whole thing divided by [HCl]. But I learned somewhere else that you are supposed to leave out HCl in the equilibrium constant, because HCl is a strong acid and thus completely dissociates. Could you explain which is right in detail and reason behind it? Thanks. Also, around 8: 50, the instructor explains that the equilibrium lies to the left side, which contains CH3COOH because "the acetic acid is not very good at donating the proton." How do we know if it is not very willing to donate a proton?(4 votes)
- Whats the relationship between Ka and pH?(1 vote)
- When the electrons from water are donated to the hydrogen, is it wrong to think that the hydrogen is attracted to lone pair? In the last 2 videos, the arrow has gone from the water to the hydrogen but is it incorrect to have the arrow going in the opposite direction?(2 votes)
- It is incorrect because the arrow shows the movement of electrons. If you draw from H+ to the lone pairs, it is wrong because it means that the electron is going to the lone pair. That is not happening since the electron Hydrogen originally had stays with the atom it was bonded with.(3 votes)
- If H2O is present in a given equation will it ALWAYS be the BLB?(1 vote)
- Nope! Water can actually be a BLB or a BLA, it is "Amphoteric"(4 votes)
- [Voiceover] Let's look at this acid base reaction. So water is gonna function as a base that's gonna take a proton off of a generic acid HA. So lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the A. Oxygen, oxygen is now bonded to three hydrogens. So it picked up a proton. That's gonna give this oxygen a plus one formal charge and we can follow those electrons. So these two electrons in red here are gonna pick up this proton forming this bond. So we make hydronium H30 plus and these electrons in green right here are going to come off onto the A to make A minus. Let's go ahead and draw that in. So we're gonna make A minus. Let me draw these electrons in green and give this a negative charge like that. Let's analyze what happened. HA donated a proton so this is our Bronsted-Lowry acid. Once HA donates a proton, we're left with the conjugate base which is A minus. Water, H2O accepted a proton, so this is our Bronsted-Lowry base and then once H2O accepts a proton, we turn into hydronium H3O plus. So this is the conjugate acid. So H3O plus, the conjugate acid and then A minus would be a base. If you think about the reverse reaction, H3O plus donating a proton to A minus then you would get back H2O and HA. Once this reaction reaches equilibrium, we can write an equilibrium expression and we're gonna consider the stuff on the left to be the reactants. We're gonna think about the forward reaction and the stuff on the right to be the products. Let's write our equilibrium expression. And so we write our equilibrium constant and now we're gonna write KA which we call the acid, the acid ionization constant. So this is the acid ionization constant or you might hear acid dissociation constant, so acid dissociation. So either one is fine. All right and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants. Over here for our products we have H3O plus, so let's write the concentration of hydronium H3O plus times the concentration of A minus, so times the concentration of A minus. All over the concentration of our reactant, so we have HA over here, so we have HA. So we could write that in and then for water, we leave water out of our equilibrium expression. It's a pure liquid. Its concentration doesn't change and so we leave, we leave H2O out of our equilibrium expression. All right, so let's use this idea of writing an ionization constant and let's apply this to a strong acid. HCL is gonna function as a Bronsted-Lowry acid and donate a proton to water which is going to be our Bronsted-Lowry base. And so we could think about a loan pair of electrons in the auction taking our proton, leaving those electrons behind. And so the auction is now bonded to three hydrogens because it picked up a proton, giving this a plus one charge. Once again let's follow those electrons in red. This electron pair picks up this proton to form this bond, so we form H3O plus or hydronium. And these electrons in green move off onto the chlorine, so let's show that. We form the chloride anion. Let me go ahead and draw in the electrons in green and let me go ahead and write a negative one charge here like that. Another way to represent this acid base reaction would just be to write out H2O plus HCL, gives us H3O plus, plus CL minus. So this is just a faster way of doing it and HCL is a strong acid. Strong acids donate protons very easily and so we can say this process occurs 100%. So we get 100% ionization. The equilibrium is so far to the right that I just drew this one arrow down over here. We get approximately 100% ionization, so everything turns into our products here and let's go ahead and write our equilibrium expression. So KA is equal to a concentration of H3O plus. So concentration of our products times concentration of CL minus, all over, right, we have HCL and we leave out water. If we think about approximately 100% ionization, we have all products here. So we have a very, very large number in the numerator and extremely small number in the denominator. If you think about what that does for your KA, that's gonna give you an extremely high value for your KA. All right, so KA is much, much, much greater than one here. That's how we recognize a strong acid. An acid ionization constant that's much, much greater than one. Now let's think about the conjugate base. All right, so let's go back up here. So we had a HCL and CL minus as our conjugate acid base pair and the stronger the acid, the weaker the conjugate base. All right, so HCL is a strong acid, so CL minus is a weak conjugate base. So let me write that here. The stronger the acid, so stronger the acid, weaker the conjugate, weaker the conjugate base. And one way to think about that is if I look at this reaction, we can think about competing base strength. All right, so here we have Bronsted-Lowry. Base water is acting as a Bronsted-Lowry base and accepting a proton. And over here if you think about the reverse reaction, the chloride anion would be trying to pick up a proton from hydronium for the reverse reaction here but since HCL is so good at donating protons, that means that the chloride anion is not very good at accepting them. So the stronger the acid, the weaker the conjugate base. Water is a much stronger base than the chloride anion. Finally let's look at acetic acids. Acetic acid is going to be our Bronsted-Lowry acid and this is going to be the acidic proton. Water is gonna function as a Bronsted-Lowry base and a lone pair of electrons in the auction is going to take this acidic proton, leaving these electrons behind on the oxygen. So let's go ahead and draw our products. We would form the acetate anions. Let me go ahead and draw in the acetate anion so negative one charge on the oxygen. Let me show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form a hydronium. All right, so H3O plus, so let me go ahead and draw in hydronium. So plus one formal charge on the oxygen and let's show those electrons in red. All right, so this electron pair picks up the acidic proton. Forming this bond that we get H3O plus. So another way to write this acid base reaction would be just to write acetic acid, CH3, COOH plus H2O gives us the acetate anion, CH3COO minus plus H3O plus. Now acetic acid is a weak acid and weak acids don't donate protons very well. So acetic acid is gonna stay mostly protonated. When you think about this reaction coming to an equilibrium, you're gonna have a relatively high concentration of your reactants here. When we write the equilibrium expression, write KA is equal to the concentration of your product so CH3COO minus times the concentration of H3O plus, all over the concentration of acetic acid because we leave water out. So all over the concentration of acetic acid. All right, the equilibrium lies to the left because acetic acid is not good at donating this proton. So we're going to get a very large number for the denominator, for this concentration so this is a very large number and a very small number for the numerator. All right, so this is a very small number. So if you think about what that does to the KA, all right, a very small number divided by a very large number, this gives you a KA value, an ionization constant much less than one. All right, so this value is going to be much less than one and that's how we recognize, that's one way to recognize a weak acid. Look at the KA value.