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

Sulfuric acid (H₂SO₄) is unique as the only strong diprotic acid, meaning it fully dissociates its first acidic proton in water, forming HSO₄^- and H₃O⁺. The second acidic proton dissociates only partially, making it a weak acid in this step, indicated by reversible arrows in the chemical equation. This two-step dissociation process is crucial for understanding the overall concentration of H⁺ ions in a solution, which is essential for calculating the pH. The constants K₁ and K₂ represent the dissociation constants for the first and second protons, respectively, with K₁ being a measure of the strong acid dissociation and K₂ representing the weaker second dissociation. Understanding these concepts is vital for comprehending acid behavior in aqueous solutions.

Sulfuric Acid

When it comes to diprotic acids, just remember that sulfuric acid, which is H2SO4, represents the only strong diprotic acid. All others will be considered weak. Here we're going to say in terms of acidity, the first acidic proton associates completely and the second acidic proton only partially. Taking into account both acidic protons will help give us a complete picture of the concentration of H+ ions.

With that information we can calculate pH. So just remember how would this look? We have sulfuric acid here reacting with water. Since it's strong, we'd have a solid arrow going forward. This would create HSO4 minus aqueous plus H3O plus. We've lost the first acidic hydrogen, so that's K1. But now if we're trying to lose the 2nd acidic hydrogen, that's HSO4 minus aqueous plus H2O liquid.

At this point it's considered weak. So instead of having a solid error going forward, we have reversible arrows, meaning only a portion of the reactants breakdown to give us products. Here this would deal with K2 where we're losing the 2nd acidic hydrogen. We create our sulfate ion plus additional H3O plus ion. So this is the picture you have to take. When it comes to sulfuric acid, the first acidic proton is lost completely, but only the second one partially.

Diprotic Acids and Bases Calculations Example

Calculate the pH of a .0550 molar solution of sulfuric acid. Here we're given the K₁ and the K₂ for sulfuric acid. Now notice that K₁ is a pretty large number, a number much greater than one. That's because the first proton of sulfuric acid is of a strong diprotic acid. K₂, though, is a number smaller than one. That's because removing the second H⁺ ion is much more difficult. It's a weaker version of the original strong diprotic acid.

With this information, let's get to it. TE one we're going to use the Bronsted-Lowry definition to predict the product's formed when H₂SO₄ reacts with water. So here we have H₂SO₄ aqueous plus H₂O liquid and here we're going to say that it's a strong asset. So it reacts completely with the water donating an entire H⁺ ion to the water molecule to create HSO₄^- aqueous plus H₃O⁺ aqueous. Now the initial concentration here of the sulfuric acid is .0550 molar. We're going to say sulfuric acid completely dissociates initially, meaning the concentration of sulfuric acid equals the concentration of hydronium ion and the intermediate form. So these would also have the same initial concentration.

With this information we can go to step 2. IN step two we set up an ICE chart for the intermediate form created in step one and reacted also with water. So the intermediate form formed in step one was hydrogen sulfate, so HSO₄^-. This is also called bisulfate. So here it reacts with water as an acid. Donating an H⁺ to the water molecule it becomes the sulfate ion and water gaining in H⁺ becomes the hydronium ion. Here we have is the initial concentration .0550, but also for the hydronium ion. Water is a liquid. Liquids and solids are ignored in an ICE chart. We don't have any information on the basic form.

Now here, using the initial row, we're going to lace the amounts from step one for the intermediate form and the H₃O⁺, which we did, we place a zero for the basic form. We now for step four, we're going to say we lose reactants to make products and we're going to say here that using the change row you're going to place A -, X for the reactants and A + X for the products. With this information, we just bring down everything that we have and that'll give us our equilibrium run. OK, so bringing down everything, this would be oh .0550 -, X + X + X. Now our ICE chart is filled in. We can say using the equilibrium row, set up the equilibrium constant expression with K_a2 and solve for X.

Now we're using K_a2 because now we're talking about losing this last and final H⁺ of the diprotic species to create our basic form. At the end here we're going to say K_a2 equals products over reactants. So sulfate ion times hydronium ion divided by the bisulfate ion K_a2. When we plug that number in, K_a2 here would equal 1.2 * 10^-2. At equilibrium, both of these products are X, so that's X² divided by 0.0. And actually, it wouldn't just be actually oh .0550 + X, so that would actually be oh .0550 + X * X / .0550 - X.

All right. So as we can see, this is going to be a little bit tricky in terms of solving this expression. We're going to have to set up a quadratic formula in order to isolate X here. Now here. Once we do that, we're going to solve for pH by adding the hydronium ion concentrations that we found in parts one and five to determine the complete H⁺ concentration or total H⁺ concentration. All right. So let's get to it. We're going to cross multiply these two together. So basically this distributes here and this distributes here. So when we do that, what are we going to get initially? When we do that, we're going to get initially 6.6 * 10^-4 -, 1.2 * 10^-2 X equals this X going to distribute here and distribute here. So make sure that's going to give me 0.055. Oh X + X².

We have all these X variables, but if this X that has the highest exponent of two, so it's our lead term, that means everything has to be moved over to the right side. We're going to add this to both sides here so it adds together with this X, and then we're going to subtract this from both sides. So when we do that, we're going to get X² + .067 X minus 6.6 * 10^-4. This will be my a, my B and my C and with this we can set up the quadratic formula. So still not done. So remember the quadratic formula itself is negative b ± sqrt b² -, 4 AC over 2A. Let's start lugging in the values that we have, right? So here we said that B is going to be negative 0.067 plus or minus .067² - 4 * 1. Don't forget the negative sign of C, so -6.6 * 10^-4 / 2 * 1.

We're going to solve for everything in here and then take the square root of it. So when we do that, we're going to get 0.067 plus or minus oh .08443 / 2. Because this is plus or minus, that means there's two possible values for X1. Value is if we added .08443 which would give me initially .008715 or we could do minus 0.08443 which would give me negative 075715. Remember the correct X is the one that where you can place it anywhere on the equilibrium row and it gives you back a positive value. The -1 will not work because if you place it here it will give you a negative concentration at equilibrium which is not possible. That means that we cannot utilize this one.

So now we're going to say what is our total H⁺ concentration. So remember H⁺ is the same thing as H₃O⁺. It would be equal to this. This is the amount that we had initially from the first setup, step one. And then we just found out what X is. The additional H₃O⁺ that was created O. We're going to say .0550 + X. So that's .0550 plus the X that we found of .008715. When we add those together, we're going to get 0.063715. That is my total amount of H₃O⁺. With that we can find pH which is negative log of H₃O⁺. So plug that in and with this information we can finally find our pH, which comes out to be 1.20. So this would be the final pH of my sulfuric acid solution.

Diprotic Acids and Bases Calculations Example

When it comes to calculating the H of weak diprotic acids, it's important to remember to utilize only KA1 to calculate the pH of the acidic form of a weak diprotic acid. Here in this question it says we have to calculate the pH of 0.115 molar carbonic acid solution. Here we're given its K1 and its K2 values. So step one is we're going to set up an ice chart for the weak diprotic acid that has a reacting with water. We're going to use the Bronston Lori definition to predict the products formed here.

This carbonic acid is going to react with water, which is in its liquid phase. Since it is the acid, it's going to donate an H to the water which acts as a base. This would create the bicarbonate ion or hydrogen carbonate ion, which is HCO3-. Water gaining an H plus becomes H3O+. This is a weak diprotic acid, so we're utilizing an ice chart which stands for initial change equilibrium. Here we're going to follow Step 2.

Next, using the initial row, place the amount given for the weak diprotic acid, so 0.115 molar. Place a zero for any substance not given an initial amount. Well, here water is a liquid. Liquids and solids within Ice Charts are ignored. The products initially are zero. We're going to say here we lose step three. We're going to lose reactants to make products. So here we're going to say using the change row, place A -, X for the reactants and A + X for the products minus X. Since we're losing reactants and plus sections, we're making products. Bring down everything. So .115 -, X + X + X.

With our ice chart filled in, we go to Step 4. Using the equilibrium row, set up the equilibrium constant expression with KA1 and solve for X. So here KA1 equals products overreactants, so it equal bicarbonate ion times the hydronium ion divided by carbonic acid. With this equilibrium expression set up, we can place numbers within the spaces for K1 and the different concentrations. And what's important here is we can check to see if a shortcut can be utilized to avoid the quadratic formula.

To do this, we utilize what's called the 500 approximation method, where we take the initial concentration of our weak diprotic acid and we divide it by KA1. If we get a value greater than 500, we can ignore the minus X within our equilibrium expression. So here we have .115 molar divided by KA1. When we punch that into our calculators, we get 267,441.9, so we got a number much greater than 500, which means within our equilibrium expression we can ignore the minus X, and by doing that we can avoid the quadratic formula altogether.

So we come back up. Here K1 is 4.3×10-7. Both products are X, so that's X2 / .115 cross multiply X2=4.945×10-8. Take the square root of both sides X=2.224×10-4, so we just found out what X is. This takes us to Step 5. Ince we're looking at the acidic form of a weak dichotic acid. The X variable will equal the hydronium ion concentration and with that we can figure out pH. So we're going to say here pH equals negative log of H3O plus concentration, so the negative log of 2.224×10-4. Punch that into our calculators and we'll get 3.65. That's the pH for the acidic form of this weak diprotic acid in the form of carbonic acid.

Diprotic Acids and Bases Calculations Example

So in this video we're going to learn how to calculate the concentration of the basic form of a weak diprotic acid. Now this is a special case, so pay close attention. We're going to say when given the initial concentration of our weak diprotic acid, the concentration of its basic form equals the K2 value. Now what is a question like this look like? Here in this example it says determine the concentration of the carbonate ion when given 0.115 molar of carbonic acid solution. So they're giving us the initial concentration of our weak diprotic acid. So it's acidic form and they're asking you to determine the concentration of its basic form.

Remember, for a weak diprotic acid we have the acidic form or it has both H+ ions. When it loses its first H+ ion, it becomes its intermediate form, and then when it loses that second H+, it becomes its basic form like we set up above. If they give you the initial concentration of the weak diprotic acid, then K2 equals the concentration of the basic form. So here we just simply say that the concentration of the carbonate ion equals K2, so it equals 5.6×10-11 molar. That will be our answer. Now if you don't want to continue on with this video, you can turn it off, because what we're about to do is we're about to show proof of why that is OK. But again, remember, if they give you the initial concentration of the weak diprotic acid, the concentration of the basic form is equal to K2.

For those of you who are continue on with this video, let's get started. So how do we prove this? Well, step one, we set up an ice chart to calculate the concentration of the intermediate form and the hydronium ion, and we're doing this by using K1. So the carbonic acid will react with water in its liquid phase. It donates an H+ to water to become the carbonate ion or bicarbonate ion. Sorry plus the hydronium ion. This is a nice chart. So it's initial change equilibrium. Remember with this we ignore solids and liquids. The initial concentration of my weak acid is 0.115 molar. These are zero. We lose reactants to make product. We bring down everything using the 500% approximation method. We would see that we could ignore the minus X.

If you're not familiar with this approach, make sure you go back and take a look at my video where I talk about the 500 approximation method. So here K1 equals products overreactants, so we plug in the numbers here. So K1 is 4.3×10-7 = X2 / 1 one five. When you cross multiply you see that X2 = 4.945×10-8, and when you take the square root of both sides you get X equals 2.224×10-4. Now that we've found X, we've found the concentrations of both the intermediate form, so your bicarbonate ion and the hydroxide ion.

With that information, we move on to Step 2, set U another ice chart and using the initial role, place the calculated amounts of the intermediate form and the hydronium ion concentration. So here we would place 2.224×10-4 molar. Here this is reacting with a second molecule of water creating the carbonate ion which is our basic form and we create more hydronium ion whose initial concentration would also be 2.224×10-4. Again this number is coming from the first ice chart where we calculated the equilibrium concentrations. Still a liquid, Place a zero for the concentration of the basic form, in this case carbonate ion set. Haul this up again. So this is initial change equilibrium. We're going to say here we lose reactants to make products, so usually change row. Again, we do minus X for reactants, plus X for products. Bring down everything in 2.224×10-4 + X.

So now here we're going to say, using the equilibrium row, set up the equilibrium constant expression. But now we're doing it with K2 and we're going to use it to solve for X. Now we're going to say we're going to check our shortcut, the 500 approximation method, to see if we can avoid the quadratic formula O. Here the initial concentration is not the very first initial concentration. If the initial concentration of our intermediate form, which is the bicarbonate and we're going to divide it by K2, when we do this, we're going to get 3964285.7. So by doing the initial concentration divided by our K2 in this in this case, we found a value greater than 500. But now we can ignore not only the minus X that we're used to seeing, but also plus X, because as we can see here, we had initial concentration of H3O plus, which gave us A + X here at the end.

So we'd ignore those two things. So we ignore this one, and we ignore this one. What else happens by ignoring the plus X and the minus X? What cancels out both of these two numbers because they're the same? And what do you have left at the end? You see that X equals the K2 value. Coming back up here in our eye chart, we can see X equals the concentration of the basic form of my diprotic acid. So again, it just simply equals K2. So this will be our answer. So all of its work was just proof to show that if they give us the initial concentration of the acidic form of the weak diprotic acid, then K2 is equal to the concentration of the basic form.

You don't need to do all this work. This is just to prove that OK, just remember, you see a question like this in the beginning, you can say, oh, this is a special case, The concentration of basic form is just simply equal to K2. Alright, so keep that in mind when dealing with this special circumstance, special case.

Which of the following diprotic acid would produce the most acidic solution when dissolved?

• A. 0.200 M H₂SO₃ (K_a1 = 1.6 x 10^-2 ; K_a2 = 4.6 x 10^-5)
• B. 0.200 M H­₃PO₄ (K_a1 = 7.5 x 10^-3 ; K_a2 = 6.2 x 10^-8 ; K_a3 = 4.2 x 10^-13)
• C. 0.200 M H­₂CO₃ (K_a1 = 4.3 x 10^-7 ; K_a2 = 5.6 x 10^-11)
• D. 0.200 M H­₂S (K_a1 = 8.9 x 10^-8 ; K_a2 = 1.0 x 10^-19)
• E. 0.200 M H­C₉H₇O₄ (K_a1 = 3.3 x 10^-4)

Determine the pH of 0.115 M Na₂S. Hydrosulfuric acid, H₂S, possesses K_a1 = 1.0 × 10⁻⁷ and K_a2 = 9.1 × 10⁻⁸.

• A. 7.04
• B. 10.05
• C. 7.90
• D. 13.06

Here’s what students ask on this topic:

How can the concentration of a diprotic acid be found?

To determine the concentration of a diprotic acid, which can donate two protons (H⁺ ions), you can perform a titration using a strong base, such as NaOH. Here's a simplified process:

1. Prepare a standard solution of the strong base with known concentration.
2. Take a measured volume of the diprotic acid solution whose concentration you want to find.
3. Add a few drops of an appropriate indicator to the acid solution, which will change color at the endpoint of the titration.
4. Slowly add the base to the acid while stirring until the indicator shows that neutralization has occurred. This is the endpoint, where moles of H⁺ from the acid equal moles of OH^- from the base.
5. Record the volume of base used to reach the endpoint.

Since the acid is diprotic, remember that one mole of acid will react with two moles of base. Use the stoichiometry of the reaction to calculate the concentration of the acid:

Molarity of acid = (Molarity of base × Volume of base used × 2) / Volume of acid

This equation accounts for the diprotic nature of the acid (hence the multiplication by 2). Ensure that all volumes are measured accurately to get precise results.

How do you calculate the pH of a diprotic acid?

Calculating the pH of a diprotic acid, which can donate two protons (H⁺ ions), involves a few steps. First, you need to write down the two dissociation equations:

1. $H_{2}A\to H+{\mathrm{HA}}_{-}$
2. ${\mathrm{HA}}_{-}\to H+A_2-$

Each step has its own dissociation constant, K_a1 and K_a2. To calculate the pH:

1. Determine the concentration of the acid and the values of K_a1 and K_a2.
2. If K_a1 is much larger than K_a2 (by 10³ or more), you can initially ignore the second dissociation.
3. Calculate the concentration of H⁺ from the first dissociation using the formula: $\left[H\right]$