Diprotic Acids and Bases - Online Tutor, Practice Problems & Exam Prep

Diprotic acids can donate two H⁺ ions, represented as H₂A. The first ionization constant, K_a1, is larger than K_a2, indicating it's easier to remove the first H⁺. Diprotic bases, represented as A²⁻, can accept two H⁺ ions, with K_b1 and K_b2 as their respective constants. The relationships between these constants are K_a1 × K_b2 = K_w and K_a2 × K_b1 = K_w, where K_w is the ion product constant for water.

Diprotic Acids

So diprotic acids and bases are compounds that can donate or accept 2 H⁺ ions. Now for diprotic acid, we're going to see that their basic formula, their generic formula is H₂A. Their equations are illustrated by the following two examples. So here, when we're dealing with the acid in its fully protonated form, meaning it has both of its acidic hydrogens, it's going to be the acid. Water is going to be our base.

Remember, following theory, the acid donates an H⁺ over to the base. As a result of this, we create A^- and H₃O⁺ as our products. Since we're talking about removing the first acidic hydrogen from our diprotic acid, that means we're dealing with K_a1. Just like all other K_a values, it equals products over reactants. Our equilibrium expression will come down to A^- times H₃O⁺ divided by H₂A.

Now here, this acid, this form that we just created, which is our conjugate base form because it lost an H⁺, could continue to lose its second H⁺ and give us our second equation. This A^- continues onward where it still acts as an acid. Another water molecule acts as the base. Just like before, the acid donates an H⁺ to the base. Now we create A^2- plus H₃O⁺.

Since we're talking about removing the second acidic hydrogen from our diprotic acid, now it is K_a2 that we're dealing with. Here, we're going to say that this is A^2- times H₃O⁺ divided by A^-. Again, we ignore water because in an equilibrium expression, we ignore solids and liquids. These are the two basic equations dealing with the diprotic acid. Before we move on to diprotic bases, just remember in terms of these equilibrium constants, K_a1 and K_a2, K_a1 is always significantly larger than K_a2.

That just means that it's always easiest to remove the first acidic hydrogen from an acid, and it's much more difficult to remove the next one. Remember, K_a1 will always be a value much greater than K_a2. Click on to the next video and continue our discussion with diprotic bases.

Diprotic Bases

Now that we've seen the diprotic acid form of a compound, we can examine its diprotic base form as well. We're going to say for diprotic bases, they can be represented by a 2 minus. The 2 minus representing the fact that it can gain 2 H⁺ ions. Here, the equations can be illustrated by the two examples given below. In this form, we have our base form.

It hasn't accepted an H⁺ yet. In this first equation, it's the base. Water is the acid. Water will donate an H⁺ to it, thereby creating HA^-. By donating an H⁺, the water itself becomes OH^-.

And because we're talking about accepting the first H⁺ion, that means we're dealing with K_b1. K_b1 is an equilibrium constant and, like all other equilibrium constants, it's equal to products over reactants. Remember, we have to ignore solids and liquids. So here it would be HA^- times OH^- divided by A^2-. Now the HA^- that we've created here can continue onward because it's still negative.

It can still accept another H⁺. We're gonna say here, we bring it down. It's still gonna act as the base. It's gonna react with a second water molecule which will act as the acid. Again, the water will donate an H⁺ to the HA^-, creating H₂A.

Water again loses an H⁺ so it becomes OH^-. Since we're talking about accepting the second acidic hydrogen, that means we're dealing with K_b2. Just like the other K_b, it's equal to products over reactants and again we ignore liquids and solids. So here this would be H₂A times OH^- divided by HA^-. So at this point, we've seen the diprotic acid form, the dibasic form, diprotic base form.

We've seen K_a1, K_a2, K_b1, and K_b2. But how exactly are they connected to one another? By examining the next section, you'll be able to see the connections that we have with these 4 equilibrium constants as well as the different forms of our diprotic compound. So click on the next video and see how we relate them to one another.

Diprotic Dissociation Constants

At this point, we've seen the diprotic acid as well as the diprotic base in action. We've seen how we deal with 4 different equilibrium constants. Now, how exactly do we relate all these things with one another? Well, we're going to say based on these equations, the relationship between the different forms of diprotic species are, we can start out with the fully protonated or acidic form of our diprotic species. That would be H₂A.

In its acid form, it only has one choice. It can choose to donate an H⁺ away. By giving away that first H⁺, that gives us HA^-. And since we're talking about losing the first acidic H⁺ that means we're dealing with K_a1. Now, HA^- which represents our intermediate form, can also decide to donate an H⁺ away giving us A^2-.

Since we're talking about removing the second acidic hydrogen, that means we're dealing with K_a2. But remember, these diprotic species, we can also look at them in terms of a base. Here for this basic species here, we could say that it accepts its first H⁺ to become HA^-. By accepting that first acidic hydrogen, that means we're dealing with K_b1. Then the intermediate HA^- can decide to accept another H⁺ to become H₂A.

By accepting that second acidic hydrogen, that means we're dealing with K_b2. Now notice, we have here K_a1 and K_b2 overlapping one another and what do they do? They relate the acid form to the intermediate form. Then we also have a K_a2 and K_b1 overlapping one another and they relate the basic form with the intermediate form. Because of this, we can establish relationships between K_a and K_b.

So, because of the overlapping, we can say here that K_a1 times K_b2 gives us K_w. We can also say because of the overlapping, K_a2 times K_b1 equals K_w. Remember, K_w represents the dissociation constant for water which is 1.0 times 10^-14 when the temperature is 25 degrees Celsius. Remember, all equilibrium constants are temperature dependent.

If we change the temperature, K_w would change. If they did that, they would tell you what the new K_w would be. Because of this, we can establish some guidelines when dealing with one of these forms of our diprotic species. So if we take a look, we're gonna say here, if we're dealing with H₂A, this is the fully protonated acidic form. Treat it as you would a monoprotic acid.

So we'd say here that we would use K_a1 when dealing with it in order to find pH. We'll skip the intermediate form for now and come back to it. A^2- has no H⁺ on it so it only has one choice, to accept an H⁺ and act like a base. Because we're accepting our first H⁺, that means that it's going to act as a base and use K_b1. Now the intermediate form is tricky because the intermediate form here can do one of 2 things.

It could either donate another H⁺ and become the basic form and in that case would use K_a2 or it could decide to gain another H⁺ to become the acid form and therefore use K_b2. So for this one, it could use K_a2 or K_b2. So it really depends on the species itself. Now realize this intermediate form has a hydrogen so it can act as an acid and a base because it's a negative charge there. If you can act as an acid or a base, remember you're called an amphoteric species.

Now amphoteric species, we have acidic ones and basic ones. The acidic ones would donate H⁺ and therefore use K_a2. Here we'd have hydrogen sulfate. We'd have hydrogen sulfite or sulfate. We'd have hydrogen sulfite or bisulfate and bisulfite.

Then we'd have here, dihydrogen phosphate. These are your most common amphoteric acidic compounds that would use your, K_a2. And technically here, this one fits into polyprotic because it has more than 1. We'll talk about that later on in our polyprotic acids video. Then we have our basic ones.

So these basic ones here would use K_b2. So who fits in here? We have hydrogen carbonate, also called bicarbonate. We have hydrogen sulfide and then we have hydrogen phosphate. Again, this falls into the arena of polyprotic species.

So for right now, don't worry about these 2 here. Just realize later on when we talk about amphoteric species that they themselves represent acidic, amphoteric, and basic amphoteric species. So again, remember when it comes to the intermediate form, it could be K_a2 or K_b2 depending on if it wants to act as an acid or as a base respectively. In terms of this topic, when we're talking about diprotic species, remember the first 2 that I listed for acidic and the first two I listed for basic. Those are the categories they fall into and determine if they use K_a2 or K_b2.

Hopefully, you guys were able to follow along in terms of this. Now that we've gotten out of the way all the theory and ideas behind diprotic species, we'll continue onward with calculations in terms of finding pH and concentrations of these different types of compounds.

Diprotic Acids and Bases Calculations 1

Here we're told that sulfuric acid, which is H₂SO₃, represents a diprotic acid with a K_a1 equal to 1.6 × 10^-2. K_a2 itself equals 4.6 × 10^-5. Now it says, calculate the pH and concentrations of sulfuric acid, hydrogen sulfide also known as bisulfite, and sulfate ion when given 0.200 molar of sulfuric acid. For sulfuric acid, we're dealing with the acidic form or the fully protonated form of the compound. Remember, when we're dealing with the acidic form of the compound, we use K_a1 to help us determine what the pH will be.

Now our diprotic acid is H₂SO₃. It will react with water. Now remember, the acid is going to donate an H⁺ to the water. It's going to become HSO₃^- Remember, charged species are aqueous when in solution and water itself becomes H₃O⁺ We have initial change equilibrium to represent the ICE chart that we've set up. Remember, in an ICE chart, we ignore solids and liquids.

Water, which is a liquid will be ignored. Now initially, we have 0.200 molar of my diprotic acid. The products initially are 0. We lose reactants to make products. Bring down everything.

At this point again, we're dealing with removing the first acidic hydrogen from our compound. So we're dealing with K_a1. K_a1 equals products over reactants. So it's HSO₃^- times H₃O⁺ divided by H₂SO₃.

We're going to plug in the values that we know, so we have 1.6 × 10^-2 equals. At equilibrium, both my products are equal to x and since they're multiplying each other, that's x squared divided by 0.200 minus x. At this point, we have to determine can we ignore the minus x or not. To do that, we do the 5% approximation method. Now with this 5% approximation method, we take the initial concentration of our diprotic species here and we divide it by the K value that we're using.

In this case K_a1. If the ratio is greater than 500, then I can ignore that minus x and avoid the quadratic formula. So our initial concentration is 0.200 molar divided by our K_a1 which is 1.6 × 10^-2. When we do that, we get back 12.5. So we have a value here that is not greater than 500, so we cannot ignore the minus x and we're going to have to do the quadratic formula.

So we're going to multiply both sides here by 0.200 minus x. Here, we're going to distribute, Distribute. So here, that's going to give me 0.0032 - 1.6 × 10^-2x equals x squared. Here, this x squared has the highest power, so it's going to be our lead term which means everything gets moved over to its side. So we are going to add 1.6 × 10^-2x to both sides and subtract 0.0032 from both sides.

When we do that, we're going to get the expression x² + 1.6 × 10^-2x - 0.0032. So this is my a, my b and my c. My quadratic formula will be negative b plus or minus square root of b squared minus 4ac over 2a. When we plug in the values, so negative 1.6 × 10^-2, plus or minus sqrt((1.6 × 10^-2)² - 4 × 1 × -0.0032) divided by 2 × 1.

So here, when we solve for everything in here and then take the square root of that, what we'll get at this point is x equals -1.6 × 10^-2 plus or minus 0.114263 divided by 2. Realize here that we have plus or minus which means we're going to get 2 possible x variable answers. So x equals 0.049131 if we added or x equals -0.065132 if we use the minus instead. How do we know which one to use? Well, at equilibrium, you cannot have a negative concentration and it's not possible.

If we use that negative x and plugged it into here or into here for equilibrium, that give us a negative value. That would mean that that is not the correct concentration to use. This one does not work. My x will be this number here. All we say now at this point is at equilibrium, H₂SO₃ equals 0.200 minus x.

Plug in what we just found in for x and what we get for equilibrium amount for sulfuric acid at equilibrium is 0.151 molar. At equilibrium, hydrogen sulfide or bisulfide equals H₃O⁺ because they both equal x, which again we said is equal to 0.049131 molar. And because we know what H₃O⁺ is, we know what pH is because pH equals -log of H₃O⁺. Here, that gives me 1.31 as my pH once I plug that in. Now, up to this point, what have we determined?

Well, we figured out what the pH is of my diprotic acid. We determined what the final equilibrium amount of sulfurous acid is and at this point, hydrogen sulfide or bisulfide.

Diprotic Acids and Bases Calculations 1

Here it states, determine the pH of a 0.080 molar sodium sulfide solution. Here we're told that hydrosulfuric acid, which is H₂S, contains k₁ equal to 1.0 times 10^-7 and k₂ equal to 9.1 times 10^-8. Now realize here that we're dealing with sodium sulfide. That's the compound we have to work with. The only reason we're given information on hydrosulfuric acid is so that we can use its K_a1 and K_a2 values.

Realize here that sodium sulfide has had all its hydrogens removed. Sodium sulfide represents the basic form of our diprotic acid. Here, it would represent 2 sodiums and then the sulfide ion. The sulfide ion represents our basic form of our diprotic acid. We can ignore the sodium because it's just a spectator ion.

We're going to bring down the sulfide ion that we have. It's going to react with water. Here it's the base, so water is going to be the acid. The acid will donate an H⁺ over to the base to give us HS^- plus OH^-. We have initial change equilibrium.

Remember, in an ICE chart we ignore solids and liquids, so we're going to ignore the liquid water. Our initial concentration is 0.080 molar. These are initially 0. We're going to lose our reactant, so minus x, to help build up our products, plus x, plus x. Bring down everything: 0.080-x, +x, +x.

Now realize here we're talking about accepting the first acidic hydrogen by the sulfide ion. Because we're accepting the 1st acidic hydrogen, that means we're dealing with K_b1. But here what do we have? We have K_a1 and K_a2. What's the connection that K_b1 has?

Well, remember, earlier we said that K_b1 and K_a2 are connected together. And when you multiply them together, that gives us K_w. So, K_b1 equals K_w divided by K_a2.

K b1 = 1.0 × 10 - 14 , 9.1 × 10 - 8

When we do that, that's going to give me my K_b1 value as 1.1 times 10^-7. That's our K_b1 that we'll be utilizing.

We're going to come over here. So K_b1, just like other equilibrium constants, equals products over reactants.

Now, we're going to input the values we know for each. So 1.1 times 10^-7 equals x² divided by 0.080 minus x. We have to check to see if I can ignore that minus x or not by doing a 5% approximation method? So we're going to take the initial concentration that we're dealing with divided by the equilibrium constant that we're utilizing, which is K_b1, and look to see if that ratio is greater than 500. Here's our initial concentration divided by the K_b1 that we're using.

That ratio is definitely greater than 500. That means that this minus x is not significant enough to do any real change, so we ignore it. All right, now we just have to solve for x right now. We're going to multiply both sides by .080.

So here x² equals 8.8 times 10^-9. Take the square root of both sides here. That gives me x equals 9.4 times 10^-5. Now, anytime you find x, it's going to give you either H₃O⁺ or OH^-. This x that we isolated helps to give us OH^-.

Because we know the concentration of our hydroxide ion, we can determine pOH, because pOH is just the negative log of hydroxide ion. So that equals 4.03. And because we know that, pH equals 14-4.03, so that's going to be 9.97 as our pH.

The key to solving these types of diprotic species questions is to determine: Are you dealing with the acid form? Are you dealing with the acidic form? Are you dealing with the intermediate form? In the two examples on this page, we've seen what to do when asked to determine the pH of the acid form and the basic form. Later on, we'll see what to do when dealing with the intermediate form of a diprotic species.

Diprotic Acids and Bases Calculations 2

If \( k_1 \) equals \( 4.46 \times 10^{-7} \) and \( k_2 \) equals \( 4.69 \times 10^{-11} \) for carbonic acid, what is the pH for a 0.15 molar solution of sodium bicarbonate? Now, realize here that they're giving us information on carbonic acid. But they're asking us to determine the pH of sodium bicarbonate. Our task is to determine which form of a diprotic species we have with sodium bicarbonate. Realize here that carbonic acid is the acid form where it has both of its acidic hydrogens.

Remember, when it loses its first acidic hydrogen, it becomes \( \text{HCO}_3^- \). So it becomes bicarbonate. Then it can lose its last acidic hydrogen to become the carbonate ion. With sodium bicarbonate, which is made up of \( \text{Na}^+ \) and bicarbonate here, we can see that we're dealing with the intermediate form of our diprotic species. Now because we're dealing with the intermediate form, our task of determining pH is really quite easy as long as we remember this particular formula.

To figure out the \( \text{H}^+ \) concentration of the intermediate for a diprotic species, we say that \( \text{H}^+ \), its concentration is equal to the square root of \( k_{a1} \times k_{a2} \times \) the formal concentration which is this 0.15 molar plus \( k_{a1} \times k_w \) which is the dissociation constant for water divided by \( k_{a1} + \) the formal concentration of our species. Alright. So all we do here is we plug in our numbers. So \( k_1 = 4.46 \times 10^{-7} \times k_2 = 4.69 \times 10^{-11} \). Our formal concentration of 0.15 molar plus \( k_{a1} \times k_w = 1.0 \times 10^{-14} \) divided by 0.46 times 10^(-7) plus 0.15 molarity.

Now when we plug all that in into our calculator, what we're gonna get is the square root of \( 2.09471 \times 10^{-17} \). Taking the square root of that gives us, for the concentration of \( \text{H}^+ \), as \( 4.6 \times 10^{-9} \) molar. So this represents the concentration of \( \text{H}^+ \). And realize if we know the concentration of \( \text{H}^+ \), then we know what the pH is because pH equals negative log of \( \text{H}^+ \). So we're going to take the negative log of that number and that gives me a pH of 8.34.

Now this number makes sense because, remember, for the intermediates, these amphoteric species which can act as acids or bases because of the presence of the hydrogen and negative charge. Remember, the basic amphoteric species, which we've said in earlier videos are bicarbonate, your hydrogen sulfide ion, and hydrogen phosphate ion. So it makes sense that bicarbonate has a pH greater than 7 because it represents one of the basic amphoteric ions. Just remember, when dealing with diprotic species especially the intermediate form, all we need to do is utilize this formula to determine the concentration of \( \text{H}^+ \). From that, we can determine pH and if we want, we can determine hydroxide ion concentration and pOH.

Just keep in mind the formula and you'll be able to solve questions just like this one.

An unknown diprotic acid has an initial concentration of 0.025 M. What is the pH of the solution if pka₁ is  3.25 and pKa₂ is 6.82?

• A. 4.21
• B. 2.46
• C. 1.91
• D. 3.25
• E. 2.79

Here’s what students ask on this topic:

What are diprotic acids and how do they differ from monoprotic acids?

Diprotic acids are acids that can donate two H⁺ ions, represented as H₂A. They undergo two ionization steps, each with its own ionization constant (K_a1 and K_a2). The first ionization constant, K_a1, is always larger than K_a2, indicating it is easier to remove the first H⁺ ion. In contrast, monoprotic acids can only donate one H⁺ ion and have a single ionization constant (K_a).

How do you calculate the pH of a diprotic acid solution?

To calculate the pH of a diprotic acid solution, you need to consider both ionization steps. First, use the first ionization constant (K_a1) to find the concentration of H₃O⁺ produced in the first step. Then, use the second ionization constant (K_a2) to find the additional H₃O⁺ produced in the second step. The total [H₃O⁺] is the sum of the contributions from both steps. Finally, calculate the pH using the formula:

$\mathrm{pH}=-\log \left(H_{+}\right)$

What is the relationship between K_a and K_b for diprotic acids and bases?

The relationship between the ionization constants of diprotic acids and bases is given by the equations:

$K_1\times K_2=K_{\mathrm{w}}$

and

$K_2\times K_1=K_{\mathrm{w}}$

where K_w is the ion product constant for water (1.0 × 10^-14 at 25°C). These relationships help in understanding the equilibrium between the different forms of diprotic species.

What are the common examples of diprotic acids and bases?

Common examples of diprotic acids include sulfuric acid (H₂SO₄), carbonic acid (H₂CO₃), and oxalic acid (H₂C₂O₄). For diprotic bases, examples include carbonate ion (CO₃^2-) and oxalate ion (C₂O₄^2-). These compounds can donate or accept two H⁺ ions, respectively, and have distinct ionization constants for each step.

How do you determine the equilibrium concentrations in a diprotic acid solution?

To determine the equilibrium concentrations in a diprotic acid solution, follow these steps:

1. Write the ionization equations for both steps.
2. Set up the equilibrium expressions for K_a1 and K_a2.
3. Use the initial concentration of the diprotic acid and the ionization constants to solve for the equilibrium concentrations of all species involved.
4. Apply the ICE (Initial, Change, Equilibrium) table method to keep track of concentration changes.

This approach helps in systematically finding the concentrations of H₃O⁺, the conjugate base, and the remaining diprotic acid at equilibrium.