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

In polyprotic acids, the removal of acidic hydrogens involves distinct acid ionization constants: K1, K2, and K3. The fully protonated form behaves like a monoprotic acid, while the fully deprotonated form acts as a base. Intermediate forms require specific equations to calculate [H⁺] concentrations, utilizing overlapping constants such as K_a and K_b.

Polyprotic Acids

At this point, we've mastered diprotic acids, and it's onwards towards polyprotic acids. Now, because of the knowledge we've gained from looking at diprotic acids, we can now apply them to our understanding of polyprotic acids. Here it says, for polyprotic acids, which we'll represent by H₃A, their equations can be illustrated by the examples given below. With H₃A, we're dealing with the fully acidic form of our polyprotic or, in this case, triprotic acid. Here it's going to give away its first acidic hydrogen to water, thereby creating H₃O⁺. Because we're talking about removing the first acidic hydrogen, we're dealing with K_a1. Here, it's equal to products over the reactant, so it's:

H 2 A − H 3 O + H 3 A

This compound still has acidic hydrogens that can still donate, so it continues onwards and gives another H⁺ to a second water molecule, creating H_A2- with H₃O⁺. Again, we're dealing with K_a2 now because we're removing the second acidic hydrogen. This is:

H A2 − H 3 O + H 2 A −

Then finally, this species that's created can give away another H⁺ to a third water molecule to produce A_3- + H₃O⁺. Our equation at the end will be:

A 3 − H 3 O + H A2 −

Now, as we said earlier with diprotic species, remember, removing the first acidic hydrogen is the easiest. With each sequential hydrogen afterward, it becomes harder and harder to remove. This weak acid will make a small amount of these products because that's the easiest H⁺ to remove. But then making these products will have very little being formed, and with the third phase, even less so. Less and less products are formed because it becomes harder and harder to remove the next acidic H⁺. Just keep in mind some of the fundamentals when looking at a triprotic or a polyprotic acid form here. Now click on to the next video and see how we tackle polyprotic bases.

Polyprotic Bases

We've looked at the polyprotic acid form. Now let's look at the polyprotic base form. Here, we're dealing with a 3 minus. So we have the potential of absorbing or accepting 3 H⁺ ions. Here, we're dealing with the basic form where it has no acidic hydrogens on it. It's going to just accept an H⁺ from water thereby creating HA2- by donating an H⁺ away, water itself becomes OH^-. Now because we're talking about accepting the first acidic hydrogen, we're dealing with Kb1. Here would be 2−×OH−∣A3−. Here, we're now going to deal with HA2−. With HA2−, water is now going to donate a second H⁺ to make it into H2A- and OH^-. Here we have H2A-×OH−÷HA2−. Finally, this one that we created can react with a third water molecule to produce these new products ×OH−÷H2A−. Realize here by examining the polyprotic acid form and the polyprotic base form, we've gone through a lot of different variations of our poly species. We've gone through the fully protonated version that is acid form which could have lost an H⁺ to become H2A−. That could have continued to lose another H⁺ to become HA2− and then finally we have a 3 minus after it's lost all of its H⁺. This understanding will help us realize the connections between the different forms of these compounds and the different Ka and Kb values that are associated with them. Click on to the next video to look at the final portion when talking about polyprotic species.

Polyprotic Dissociation Constants

So now that we've looked at the polyprotic acid form as well as its base form, let's connect them all together. Now, here as we said that we have the fully protonated version. Here we have our intermediate 1 after it's lost its first acidic hydrogen. Then it's lost its second to give us intermediate 2. And then we have our basic form here. Realize to remove the first acidic hydrogen means that we have Ka1 involved. To remove the second is Ka2 and finally to remove the 3rd and last hydrogen is Ka3. Going the opposite way, adding the 1st acidic hydrogen would be Kb1. Adding the second would be Kb2 and adding the 3rd would be Kb3. Just like diprotic species, we have overlapping of equilibrium constants. We can see that Ka1 and Kb3 overlap so that's why Ka1 ⁢ Kb3 = kw. Then because they overlap finally here.

Now when dealing with polyprotic acids, we're going to say here that when we have the fully protonated version, H₃A, where it has all 3 of its acidic hydrogens, we're going to treat it as though it were a monoprotic acid and use Ka1. When we're dealing with A^3- where it's lost all of its hydrogens and all it can do is accept the first one, we're going to treat it as though it's a base and use Kb1. Then finally, if we're dealing with intermediate 1 where it's lost its first acidic hydrogen, we use this equation here to determine the concentration of H⁺. Here it's KA1 ⁢ KA2 ⁢ c + K1 ⁢ kw / (K1 + c). Recall that this is the same equation we used for the intermediate of a diprotic species. But now we're talking about it in terms of the first intermediate of a polyprotic or triprotic species. And then finally, here we have our 2nd intermediate form. And since we're talking about removing the second hydrogen to get to this form, we deal with K2 ⁢ Ka3 ⁢ c + Ka2 ⁢ kw / (Ka2 + c). Remember, when we're looking at a polyprotic or a triprotic species, we have 4 possible outcomes. When we're dealing with the fully acid form, the fully basic form and then its 2 intermediates. The intermediate forms are easier because we just have to recall the formulas, plug in the values, and find our concentrations of H⁺. With the acid form, we'd have to set up an ICE chart and use Ka1. With the base form, we'd have to set up an ICE chart and use Kb1. Just recall all of these different avenues we can take in order to determine our pH or H⁺ concentration for each one of these compounds.

Polyprotic Acid Calculations

So here it says, calculate the equilibrium concentrations of phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate, and the hydronium ion for 0.35 molar of phosphoric acid. Alright. Here we're dealing with a triprotic or polyprotic acid and that's highlighted by the fact that we have 3 K_a values given to us. Here in this triprotic or polyprotic acid, we have all 3 acidic hydrogens. Therefore, we know that it's going to start off as an acid. It will react with water. Because it's in its acidic form, we know that it's going to donate an H⁺ to water. By doing this it becomes H₂PO₄^- and water becomes H₃O⁺. Here we're going to say initial change equilibrium. We know that we're going to ignore solids and liquids. The initial concentration of this weak acid is 0.35 molar. These are initially 0. We're losing the reactant so that we can build up our products. Bring down everything. Since we're talking about removing the first acidic hydrogen or donating it to water, we know that we're dealing with K₁ . Here that's equal to our products over our reactants. Times H₃O⁺ divided by our reactant. Now we plug in the values we know for each term. So K₁ is 7.2×10-3 . Both of these are x so that's x squared on top divided by 0.35 minus x on the bottom. We use the 5% approximation method here so the initial concentration of 0.35 divided by the K_a1 value we're using. When we do that, you should see that you don't get a value greater than 500. Because of that, you'll have to keep the minus x here and perform the quadratic formula. So we're just multiplying both sides by 0.35 minus x. So distribute, distribute. When we do that, we're gonna get 0.00252 minus 7.2 × 10^-3 x equals x squared. Our lead term is the x squared since it has the highest power, So everything has to be moved over to its side. When we do that, we're gonna get x squared plus 7.2 × 10^-3 x minus 0.00252. This represents my a, my b and my c. So using the quadratic formula, we're gonna have negative 7.2 × 10^-3 plus or minus  7.2×10 -3 -4×1×-0.00252 divided by 2 times 1. Here, that's gonna give us negative 7.2 × 10^-3 plus or minus 0.100657 divided by 2. Now remember, because it's plus or minus, that means we're going to get 2 different variables, answers for x. So x can equal 0.0467 molar. Or if you do the negative approach, x equals -0.0539 molar. Realize here that at equilibrium, you cannot have a negative value. It's not possible. That means that this negative x gets dropped out. That x represents our correct answer. Now we're gonna say at equilibrium, H₃PO₄ equals 0.35 minus x. So plug in the value we just found for x. So that equals 0.3033 molar. At equilibrium, H₂PO₄^- equals H₃O⁺ because they both equal x. So their concentrations are 0.0467 molar. Alright. So, so far, we've found the concentration for our phosphoric acid, our dihydrogen phosphate, as well as our hydronium ion. We still have to figure out what our hydrogen phosphate and our phosphate ions will be. Now, we need to realize at this point, I've created this first intermediate. Because it still possesses acidic hydrogens, it can continue to donate H⁺ to other water molecules, and that's what's going to happen. We're going to write a new equation. Now that dihydrogen phosphate will react with another water molecule. It'll donate an H⁺ to that new water molecule and therefore become HP₄^2- plus H₃O⁺ being created. We have again initial change equilibrium. Again, we ignore water because it's a liquid. Initially, HPO₄^2- this is the first time we're seeing this so initially it's 0. At equilibrium in the previous ICE chart we found out that both your dihydrogen phosphate and your hydronium ion equaled x which equaled this number. So they still equal that number and now we're creating a new ICE Chart, so their initial amounts are that number. Now, we're gonna say again we're still losing reactants here in order to make these products. Bring down everything. Now we're going to say here at this point, since we're dealing with the second acidic hydrogen being removed to create our second intermediate, we're dealing with K₂ , which equals products over reactants. Coming back up here, let's look at what K_a2 is. So K₁ is 7.2×10-3 , Ka₂ is 6.3×10-8 . Look at the differences in the powers, negative three to negative eight. Remember I've said before in the past that the first acidic hydrogen is the easiest and each hydrogen afterward becomes harder and harder to remove, which is why we see a big decrease in our Ka values. They're different by a magnitude of 5. What does that mean? Well, that means that H₂PO₄^-, although it acts as an acid and donates H⁺ because its Ka₂ is so small, very, very little product is formed which means it's gonna lose a variable with x which means that this x is so small that it's not gonna really affect the final amount for H₂PO₄^-, and therefore it can be ignored. Same thing can be said for H₃O⁺. This plus x, because x is such a small number, it's not going to really increase the concentration of H₃O⁺ so it can be ignored. This x we have to keep around because we're looking for the concentration of HPO₄^2-, and that's all we have for there. So going back again, Ka 2 is 6.3×10-8 .

Polyprotic Acid Calculations 2

Here it states to determine the pH of 0.250 molar sodium hydrogen phosphate. They inform us that phosphoric acid contains Ka1, Ka2, and Ka3 with those respective values. We consider these values because they are necessary for the calculations involving sodium hydrogen phosphate. As phosphoric acid is a triprotic or polyprotic acid, its acidic form is H3PO4. Upon losing the first acidic hydrogen, it becomes H2PO4-. Losing the next hydrogen yields HPO4-2 and finally, it becomes PO4-3. In this question, we are working with sodium hydrogen phosphate which is HPO4-2, which is the form being considered.

At this stage, as we deal with "intermediate 2," we do not need ice charts or similar methods. We just use the provided formulas to determine the concentration of the hydronium ion. Specifically, the hydronium ion concentration is equal to the square root of ka2×ka3×the formal or initial concentration + ka2×kw, the dissociation constant of water, divided by ka2+formalorinitialconcentration. Plugging in the values yields: 6.210-eight, 4.210to the negative 13, 0.250 molar +6.210to the negative 8, 1.010to the -14 divided by6.210to the negative 8 +0.250 molar. Thus, it equals the square root of 2.852×10-20. Consequently, H+=1.69×10-ten, indicating the pH as the negative logarithm of this concentration.

Thus, pH=negativelog of1.69×10 to the negative 10, which calculates to approximately 9.77.

When examining such polyprotic species, identifying whether we are dealing with one of the intermediate forms is crucial in determining the path to calculate the pH of the solution. Now, armed with this understanding, proceed to example 2 and try to identify what form is being considered and the appropriate steps to find the pH.

Polyprotic Acid Calculations 1

Here it states, determine the pH of 0.150 molar citric acid. Here we're dealing with the fully protonated form of citric acid. It has all three of its acidic hydrogens. We know it has three acidic hydrogens because it possesses three Ka1 values. Because we're dealing with its first acidic form, that means we're going to deal with Ka1. The acid itself means that water will be the base. So it will donate an H+ to the water becoming H2CC6H5O7−. Water gains that H+ to become H3O+. We know we have initial change equilibrium. Remember, we ignore solids and liquids. The initial concentration is 0.150 molar. These products initially haven't formed yet, so they're 0. We lose reactants in order to make products. We bring down everything. Now at this point, we'd say that our Ka1 equals products over reactants. So x2 divided by 0.150−x. We can do our 5% approximation method to see if we can ignore that minus x, and therefore, ignore the quadratic formula. So, we do initial concentration divided by the constant that we're using, which in this case is Ka1. As long as the ratio is greater than 500, we can ignore the minus x. Now if we took that initial concentration and divide it by our Ka value, we would get approximately 202.7 which is not greater than 500. We have to keep the minus x here in our expression and do the quadratic formula. Since we're going to need room, let me take myself out of the image. Alright. So we're going to plug in our Ka1 which is 7.4×10−4 x2 over 0.150−x. So we're going to multiply both sides by 0.150−x. So this gets distributed. So we'll have 1.11×10−4−7.4×10−4x=x2. Because this x here has the largest power, it's the lead term, so everything has to be moved over to its side. So add 7.4×10−4x to both sides. Subtract 1.11×10−4 from both sides. So x2+7.4×10−4x−1.11×10−4. So this is my a, my b, and my c. Remember, your quadratic formula is −b+−b2−4ac÷2a. So let's input those values that we know. So −7.4×10−4, plus or minus 7.4×10−42−4×1×−1.11×10−4÷2×1. Remember, it's plus or minus, so we're going to get two values. So x will equal −7.4×10−4+0.021084÷2, or it equals −7.4×10−4−0.021084÷2. So my x will equal 0.010172M or −0.010912. Now remember, these x values that we find represent these x values here in our ICE chart. Remember, at equilibrium, you cannot have a negative concentration. It's impossible. Hence, we cannot use this x. This is the x that we would use. That x is equal to H3O+. pH, remember, is negative log of H3O+ or H+. Plug in that concentration we just found. That equals 1.99 as my pH. Remember, when it comes to polyprotic acids, it's always imperative to know which form you're dealing with. Knowing the form will dictate which path we take in order to find our pH.

Here’s what students ask on this topic:

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

Polyprotic acids are acids that can donate more than one proton (H⁺) per molecule. They have multiple acidic hydrogens, each with its own dissociation constant (K_a1, K_a2, K_a3, etc.). In contrast, monoprotic acids can donate only one proton per molecule and have a single dissociation constant (K_a). The dissociation of polyprotic acids occurs in steps, with each step having a distinct equilibrium constant. For example, H₃A dissociates to H₂A^-, then to HA^2-, and finally to A^3-, each step involving a different K_a value.

How do you calculate the pH of a solution containing a polyprotic acid?

To calculate the pH of a solution containing a polyprotic acid, you need to consider the stepwise dissociation of the acid. For the fully protonated form (e.g., H₃A), use the first dissociation constant (K_a1) and set up an ICE (Initial, Change, Equilibrium) chart to find [H⁺]. For intermediate forms, use specific equations that involve overlapping constants. For example, for the first intermediate (H₂A^-), use the equation involving K_a1 and K_a2. For the fully deprotonated form (A^3-), treat it as a base and use K_b1 to find [OH^-], then convert to pH.

What are the equilibrium constants K_a and K_b in the context of polyprotic acids?

In the context of polyprotic acids, K_a (acid dissociation constant) represents the equilibrium constant for the dissociation of each acidic hydrogen. For example, K_a1 is for the first dissociation, K_a2 for the second, and K_a3 for the third. K_b (base dissociation constant) represents the equilibrium constant for the acceptance of protons by the deprotonated forms. For instance, K_b1 is for the first proton acceptance by the fully deprotonated form (A^3-), K_b2 for the second, and K_b3 for the third. These constants are related by the equation K_a * K_b = K_w, where K_w is the ion-product constant of water.

How do you determine the concentration of H⁺ for intermediate forms of polyprotic acids?

To determine the concentration of H⁺ for intermediate forms of polyprotic acids, use specific equations that involve the dissociation constants and the initial concentration of the acid. For the first intermediate (H₂A^-), the equation is:

Ka1Ka2C + Ka1KwKa1 + C

where C is the initial concentration. For the second intermediate (HA^2-), the equation is:

Ka2Ka3C + Ka2KwKa2 + C

These equations help calculate [H⁺] for the intermediate forms.

What is the significance of overlapping equilibrium constants in polyprotic acids?

Overlapping equilibrium constants in polyprotic acids are significant because they simplify the calculations for certain forms of the acid. For example, K_a1 and K_b3 overlap, meaning K_a1 * K_b3 = K_w. This relationship helps in understanding the behavior of the acid and its conjugate base forms. When dealing with the fully protonated form (H₃A), you use K_a1 as if it were a monoprotic acid. For the fully deprotonated form (A^3-), you use K_b1 to treat it as a base. These overlaps allow for more straightforward calculations of pH and [H⁺] for different forms of the polyprotic acid.