Acid-Base Equilibria

Added: 2026-05-19

[00:00:00]Okay. So, let's now look at acid-base equilibria.

[00:00:10]Uh most most reactions in in chemistry uh occur in aqueous uh solutions.

[00:00:30]Okay. So, most reactions uh uh occur in aqueous solutions that we we know. Okay. So, for example, the um acid-base uh acid-base reactions acid-base reactions uh you know that this normally you you perform reactions in the lab doing some titrations. Okay. So, that occurs in aqueous solutions. We also have the buffered the buffered uh solutions.

[00:01:14]The buffered uh solutions uh which contain uh components components that enable the solution to resist a change in uh uh in pH. Okay. So, a buffer is able to resist a change in pH when hydroxide when hydroxide uh ions or or protons or protons are added.

[00:01:49]Okay.

[00:01:50]So, hydroxide ions or or protons are added. So, OH or the hydrogen ion with that is referred to as uh protons. Okay? So, a buffering a buffering uh system, or a buffered solution, is able to resist change.

[00:02:12]Uh resist change change in its in its uh pH in its uh pH. Okay? So, we know that pH uh pH can be calculated in many ways.

[00:02:29]Okay? The the the the the one that is most famous is pH is equal to negative log hydrogen concentration. So, we normally use this equation for a strong for a strong electrolyte. Okay? So, buffered uh solutions are especially important in living uh systems, which can survive only in a relatively narrow pH pH range. Okay? So, the most important uh we know that the human human body the human body the human body uh contains, or we can simply say the human body, or blood, the human blood, the human blood uh contains many buffering systems.

[00:03:28]Many buffering systems.

[00:03:31]Many buffering systems, but the most the most important of these is a mixture mixture [snorts] of carbonic the mixture of carbonic acid the mixture of carbonic acid with a concentration of 0.01 10.0 0.0012 molar and bicarbonate >> [snorts] >> and bicarbonate and bicarbonate with a concentration of that's bicarbonate ion with a concentration of 0.024 molar.

[00:04:28]Okay. So, these concentrations produce a pH a pH of of 7.4 for normal for normal blood.

[00:04:42]7.4 for normal for normal blood. Okay.

[00:04:47]So, because our cells our cells we know that our cells are so sensitive uh to pH. So, it is important that this pH is is maintained. Uh for example, uh sometimes you may consume food you may eat food that are acidic.

[00:05:07]Okay. So, we do not expect this pH to to drop. Okay. Or sometimes you may eat uh foods that contains bases that are alkaline in nature. So, we do not expect this pH uh to to increase. Okay. So, the buffering system in the body is able to resist those changes when hydroxide or or protons are uh added. Okay. Though sometimes a change may be there, but it's it's minimal. Okay. So, now uh let's look at solutions of acids solutions of acids or bases containing common ions.

[00:06:07]Solutions solutions of acids or bases containing containing common common ions. Solutions of acids or bases containing common ions.

[00:06:35]Okay. So, suppose we have a weak acid.

[00:06:39]Suppose we have a weak a weak acid. Suppose we have a weak acid, hydrofluoric.

[00:06:54]Hydrofluoric acid hydrofluoric acid. So, this is hydrofluoric acid uh which has KA it has KA of 7.2 * 10 ^ - -4.

[00:07:14]Okay. And it's salt and it's salt sodium fluoride sodium fluoride and it's salt sodium sodium fluoride.

[00:07:30]Okay. So, we have the weak acid uh HF and it's salt sodium fluoride. Okay. So, when the salt uh this is um uh a strong electrolyte. Okay. So, when when when this salt dissolves in water, it breaks in it breaks um it breaks up completely into its ions. So, we can see uh sodium sodium fluoride sodium fluoride will form sodium ion.

[00:08:14]Sodium ion plus fluoride ion. Okay. So, this one is a strong strong electrolyte.

[00:08:24]Strong electrolyte. Okay.

[00:08:28]It completely ionizes in uh in solution. Okay. So, now when we we mix um the solution of uh sodium fluoride and uh the acid we know that the acid, though it's a weak acid, it will also it will also ionize in solution to form so hydro hydrofluoric acid will ionize or dissociate in water to form hydrogen ion plus fluoride fluoride ion. Okay. So, these two are in one one vessel. So, now you can see that there's a fluoride ion which comes from the salt. There's also a fluoride ion that comes from the acid. So, this is a common ion since it is produced by both the acid and and the salt. Okay. So now, according to Le Chatelier's principle, since more fluoride more fluoride ions are are being produced, more fluoride ion are being produced, uh these when we add these fluoride ion from the acid and from the from the salt, so there'll be more fluoride ion in the in the vessel. So in this case, the equilibrium the equilibrium will shift to the to the left. The equilibrium will shift to the left. Okay. So, um we know that we use uh Le Chatelier's uh principle. Le Chatelier.

[00:10:35]Le Chatelier's principle.

[00:10:41]Okay.

[00:10:42]So, uh the the the shift the shift in equilibrium position the shift in equilibrium um position that occurs because of the addition of an ion already involved in equilibrium reaction is what we call common ion ionic effect.

[00:11:12]Common ionic uh effect.

[00:11:15]Okay. So, this effect makes a solution makes a solution of of sodium fluoride plus uh plus hydrofluoric acid less acidic.

[00:11:34]Less acidic.

[00:11:37]Less acidic than a solution.

[00:11:41]Than a solution.

[00:11:44]Than a solution of hydrofluoric acid alone. Okay. So, now we have seen that when you introduce the salt uh into this acid then the acid dissociation, the extent of dissociation of the acid will be greatly reduced by the presence of the common ion. As a result, this solution this solution will become less acidic, less acidic. Than a solution of hydrofluoric acid alone.

[00:12:34]Okay. So, that is the effect of this common common ion. Okay. So, it will make the equilibrium the equilibrium will shift away from the added component. Since more fluoride ions will be produced, so the equilibrium now will shift shift to the to the left to the left. Okay. To balance up the the fluoride. Okay. So, now the the common ion is quite general. For example, uh we can also look at ammonium chloride.

[00:13:25]Uh for example, uh, for example, ammonium chloride.

[00:13:32]Ammonium chloride, ammonium chloride, uh, added added to um to a 1.0 molar 1.0 molar ammonia solution.

[00:13:49]1.0 molar ammonia ammonia solution produces additional produces produces additional additional ammonium ions.

[00:14:06]Produces additional ammonium ammonium ions. Okay? So, this one can dissociate as follows: ammonium ammonium chloride and then plus plus water.

[00:14:25]So, it will give us ammonium ion ammonium ion plus plus chloride chloride ions. Okay? And this causes the position of the of the ammonia water water equilibrium to shift to the to the left. Okay? So, ammonia ammonia then plus water plus water then will give us ammonium ion plus ammonium ion plus hydroxide hydroxide ion. Okay?

[00:15:17]Plus hydroxide ion. So, in this case, the equilibrium will shift shift to the left. Equilibrium shifts to the uh to the left. Shifts to the to the left.

[00:15:33]Okay. So, this reduces the equilibrium concentration.

[00:15:39]This reduces reduces the equilibrium It reduces the equilibrium concentration of hydroxide uh ions of hydroxide uh ions. Okay.

[00:15:56]So, the common ion effect is also important in solutions of poly- polyprotic um polyprotic acids. Polyprotic acids. Okay.

[00:16:13]So, the production of protons by dissociation steps by the first dissociation step greatly inhibits greatly inhibits the succeeding dissociation steps.

[00:16:32]Which, of course, also produce protons.

[00:16:38]The common ion in this case. Okay. The protons, the common ion in this case.

[00:16:46]Okay. So, having done that, uh let's look at some equilibrium calculations.

[00:17:00]>> [snorts] >> Equilibrium calculations.

[00:17:10]Equilibrium calculations.

[00:17:13]So, example.

[00:17:17]Example. So, in this example, we shall look at acidic [snorts] solution.

[00:17:25]Acidic solution containing containing a common ion. Containing a common ion. Containing a common ion. Okay? So, the equilibrium The equilibrium The equilibrium concentration The equilibrium concentration of hydrogen ion The equilibrium concentration of hydrogen ion in a 1.0 molar in a 1.0 molar hydrofluoric acid hydrofluoric acid solution.

[00:18:17]Hydrofluoric acid solution is is 0.032 molar. 0.032 molar.

[00:18:32]And the percent And the percent dissociation And the percent dissociation And the percent dissociation of hydrofluoric acid and the percent dissociation of hydrofluoric acid is 3.2 3.2 percent.

[00:19:05]Then, calculate Calculate the concentration of of hydrogen ion and the percent and the percentage dissociation.

[00:19:24]And the percentage dissociation and the percentage dissociation of hydrofluoric acid of hydrofluoric acid in a solution containing in a solution containing in a solution containing 1.0 molar hydrofluoric acid. 1.0 molar hydrofluoric acid. We know KA is equal to is [snorts] equal to 7.2 times 10 to the power negative negative four.

[00:20:08]Uh and 1.0 molar and 1.0 molar and 1.0 molar sodium sodium fluoride. Okay. So we we what we need to what we need to look at here we want to compare the percentage dissociation. So the percent dissociation for for hydrofluoric acid when it is alone it is 3. 3.2. Okay. So now let's calculate the percent dissociation when it is now mixed with with sodium with sodium fluoride. It's It's salt. Okay. So to solve this question uh Okay. We can also look at the major species.

[00:21:07]the major species in solution when the two when the two are mixed. Okay? So, the major species in solution when hydrogen fluoride hydro hydro when hydrofluoric acid and sodium fluoride are mixed. So, the major species in solution is hydrofluoric acid fluoride ion.

[00:21:36]We have also sodium ion. Sodium ion and and water.

[00:21:42]And water. So, these are the major species. So, this one is a weak acid. It partially dissociates in in solution. Okay? So, now we know that this one the the sodium the sodium ion the sodium ions have neither acidic neither acidic nor nor basic nor basic uh properties.

[00:22:14]They have neither acidic nor basic uh properties. Then water water is a very weak acid.

[00:22:24]Very weak.

[00:22:26]Very weak acid. Water is a very weak acid or or base. Okay? So, these these two do not participate do not participate in the dissociation in the acid dissociation equilibrium.

[00:22:47]Okay? So, now what participates is simply these two.

[00:22:55]That's what controls the the concentration the acid dissociation equilibrium. Okay. So, we rule out the sodium and and sodium ion and and water.

[00:23:13]Okay. So, we are going to use the ICE table.

[00:23:17]ICE table. So, we we write the equation.

[00:23:23]Uh sodium hydrofluoric acid then forms hydrogen ion.

[00:23:32]Hydrogen ion plus fluoride plus fluoride ion. Okay. So, this is the the the equation. Okay. So, we write ICE ICE ICE. Okay. So, the initial concentration of the acid The initial concentration of the acid is 1.0 1.0 M molar.

[00:24:03]Then, the initial concentration of the hydrogen, that is zero. Then, the initial concentration of fluoride ion that is also 1.0 M molar. The fluoride ion that comes from the added sodium fluoride. Okay. So, change will be minus X plus X plus X. Okay. Then, the equilibrium 1.0 minus X X 1.0 minus plus X.

[00:24:44]Okay. So, this is the the table. That is how the table will look like. Then, now we write the KA expression.

[00:24:57]KA expression. So, we know that KA will equal to concentration of hydrogen ion, concentration of hydrogen ion times concentration of fluoride ion over concentration of the acid.

[00:25:16]Over concentration of the acid, which is given as which is equal to 7.2 7.2 * 10 to the power negative negative four. That is the KA given. Okay. Then now, we substitute the equilibrium concentrations into the into the expression. So, we have 7.2 7.2 * 10 to the power negative four is equal to hydrogen concentration, which is X.

[00:25:53]X then fluoride concentration, that is 1.0 + X then over over concentration of the the acid, which is 1.0 minus X. 1.0 minus X.

[00:26:13]Okay. So, to solve for X, we need now to solve for X. So, to solve for X, we are now going to to use some assumptions, what we call the 5% rule. Okay. So, we assume that X is so so small. X is so small compared to 1.0.

[00:26:36]So, therefore, we neglect X. And the the assumption is valid by 5% rule. Okay.

[00:26:44]So, if X is neglected, this equation will be 4 7.2 * 10 to the power negative four, which is equal to X.

[00:26:54]Then, it will be 1.0 over 1. 1.0. So, you can see that the 1.0 will cancel out. So, we can safely we can safely say X is equal to 7.2 * 10 to the power -4. Okay.

[00:27:15]So, approximations approximations approximations are valid are valid by 5% rule.

[00:27:30]5% rule.

[00:27:33]Okay.

[00:27:34]Approximations are valid by 5% rule. 5% rule. Okay. So, if X is equal to that, that is the concentration of of hydrogen. Okay. So, we know that concentration of hydrogen concentration of hydrogen is equal to X is equal to X, and our X has been given there. So, which is equal to 7.2 * 10 to the power -4 molar. Okay. So, this is the concentration of of X. So, if you know the concentration of X, you can as well calculate the the pH. We know that pH is equal to negative log hydrogen concentration. So, negative log negative log log 7.2 * 10 to the power -4 will give us uh it's going to give us So, you can do some more calculations there. So, the pH when you calculate, it will give you 3.114.

[00:28:51]So, that is the the pH. That is the pH of the solution. So, we have calculated the the the concentration of hydrogen. So, now what we need What we need is the percentage dissociation. So, let's calculate now the percentage dissociation. So, we we keep in mind the concentration of hydrogen because we shall use it. So, we'll write it here. Concentration of hydrogen, 7.2 * 10 ^ -4 M. Okay. So, now let's calculate the percentage dissociation.

[00:29:35]We can now calculate the percentage dissociation.

[00:29:39]Okay. So, percentage dissociation percentage dissociation is given by hydrogen ion equilibrium concentration over initial concentration of the acid, then * 100 100%.

[00:30:01]Okay. So, now we we simply substitute there, which is straightforward. So, the the equilibrium concentration of hydrogen is what we have calculated, which is 7.2 * 10 ^ -4 M.

[00:30:19]Uh Then, over initial concentration of the acid is given there, 1.0 M. Uh 1.0 M.

[00:30:27]Then, * * 100 * 100 um percentage, of which when we divide, we are going to get 0.

[00:30:39]0.072 percentage. Okay. So, this is a percentage dissociation in the presence in the presence of the common ion. So you have seen that the the percentage dissociation has greatly reduced. When the acid was alone, the percentage dissociation was 3.

[00:31:06]3.2 % but once the common ion was introduced, it has greatly reduced. So we can conclude that the common ion inhibits the common ion inhibits the acid dissociation.

[00:31:28]Inhibits the acid dissociation.

[00:31:33]Okay, so that is how we can calculate. So we can also look at some factors factors affecting percentage dissociation.

[00:31:47]Factors affecting percentage dissociation.

[00:32:01]Factors affecting affecting percentage dissociation.

[00:32:12]Factors affecting percentage dissociation. So number one is concentration.

[00:32:21]Uh concentration.

[00:32:23]Okay, so as a weak acid as a weak as a weak acid solution as a weak acid solution as a weak acid solution becomes more diluted, becomes more diluted.

[00:32:47]Becomes more diluted, the percent dissociation increases.

[00:32:53]The percent dissociation the percent dissociation increases.

[00:33:03]Percent dissociation increases.

[00:33:07]Okay? Then number two, we have also KA KA value.

[00:33:14]The KA value. Okay, so a larger KA a larger KA value a larger KA value indicates a larger KA value indicate indicate a stronger indicate a stronger weak acid.

[00:33:39]Indicate a stronger weak acid.

[00:33:45]A stronger weak acid.

[00:33:49]And the results and the results see in higher percent results in higher percent dissociation.

[00:34:03]Higher percent dissociation.

[00:34:07]Okay? [snorts] Higher percent dissociation.

[00:34:11]Okay? So these are some of the factors that affect uh percent dissociation.

[00:34:18]Okay? Another example.

[00:34:26]We are still calculating percent dissociation.

[00:34:29]Okay? So for a for a 0.100 molar solution solution of a weak acid of a weak acid with um with a 1.33 * 10 to the power * 10 to the power -3 molar molar concentration molar concentration of hydrogen ion of hydrogen ion of hydrogen ion at equilibrium.

[00:35:25]Hydrogen ion at equilibrium.

[00:35:31]Calculate uh calculate the percent dissociation. Calculate the percent dissociation dissociation. So, uh solution uh we know that percentage dissociation So, this is the the equation uh for a 0.100 molar solution of a weak weak acid with a 1.33 * 10 to the power -3 molar concentration of hydrogen ion at equilibrium. So, calculate the percentage dissociation.

[00:36:23]Calculate the percentage dissociation.

[00:36:27]So, we apply the formula percent dissociation percent dissociation, which is equal to hydrogen ion concentration over acid then times the times 100. Okay? So, the equilibrium concentration of of hydrogen it's given 1.33 1.33 * 10 ^ -3 molar then concentration of the initial concentration of the acid uh that is 0.

[00:37:05]100 molar. Okay? Then now you you divide and multiply. Then you're going to find 1.33%. Okay? So, as you can see the percent dissociation is less than less than 5%.

[00:37:28]So, it means our assumptions are uh valid. Okay? They are valid.

[00:37:35]So, percent uh percent dissociation of a weak acid measures measures the fraction.

[00:37:52]Measures the fraction.

[00:37:56]Percent dissociation of a weak acid measures the fraction.

[00:38:01]Measures the fraction.

[00:38:04]The fraction of acid the fraction of acid molecules measures the fraction of acid molecules that ionize that ionize that ionize in water.

[00:38:22]That ionize in in water. That ionize in water usually resulting usually resulting resulting in low percentages as we have seen in low percentages, less than 5%.

[00:38:48]Less than 5% for weak acids.

[00:38:53]for weak acids. Okay? For weak acids. So, that is how uh we can define or explain uh percent dissociation.

[00:39:11]Okay?

[00:39:13]So, now uh we can now look at uh buffered uh solutions or we shall look at buffered solutions in the next in the next lecture. Okay? So, the next one we shall look at buffered buffered solutions.

[00:39:40]Buffered solutions.