Fractional Compositions and Concentrations: Videos & Practice Problems

Understanding fractional compositions in acid-base chemistry is crucial for determining the concentrations of weak acids and their conjugate bases at specific pH levels. For monoprotic acids, the equilibrium expression is given by the acid dissociation constant, Ka = [H⁺][HA] . At pH = pK_a, the concentrations of the acid and conjugate base are equal. In diprotic systems, multiple forms exist, requiring consideration of both K_a1 and K_a2 for accurate calculations of each species' fraction as pH changes.

Fractional Composition

So fractional compositions are just a way of determining the amount of acid to base at a given or specific pH. As our pH decreases our solution becomes more acidic because the amount of acid present is increasing. As we increase pH, it becomes more basic because we're increasing the amount of conjugate base. Now remember with weak acids we have the formation of an equilibrium because weak acids are weak electrolytes, so they don't completely ionize into their products. Here our weak acid, which is just a simple generic formula, partially ionizes to form H⁺ ion and A^- as our conjugate base.

Setting up the equilibrium expression gives us the acid dissociation constant Ka equals the ions as products on top divided by the initial concentration of my weak acid. The formal concentration of my solution equals the concentration of my weak acid plus the concentration of my conjugate base. When we talk about the fraction of acid molecules present, we use the variable alpha to represent h a here. If we take a look here, what this chart is showing me is at a pH, the lowest possible pH, the composition of my solution will be 100% in the acid form, and 0% of the basic form. As the pH begins to increase as we move from left to right, we're going to have an increase in the amount of conjugate base and a decrease in the amount of weak acid.

Here, they intersect at this point here. At that pH, we have 50% of the weak acid form and 50% of the conjugate base form. This is true when the pH is equal to my pKa. So when those two equal each other, it's because the amount of conjugate base and weak acid are equal in amount. Remember this coincides with the whole Henderson-Hasselbalch equation, which is pH equals pKa plus log of conjugate base over weak acid.

When they're both equal to each other, this all becomes equal to 1. The log of 1 is simply equal to 0, and that's why pH equals pKa. So when pH equals pKa, the amount of weak acid is equal to the amount of its conjugate base, and if we continue past this pH point, we're going to continually have a drop in my weak acid form and a steady rise in the amount of my conjugate base form, until we get to a pH where the weak acid form is basically near 0 and the conjugate base is near 100%. Now here we'd say that the concentration of my weak acid is equal to the concentration of H⁺ times the formal concentration of my solution divided by H⁺ plus Ka. Here, if we want to find out the fraction that exists in the acid form, that just simply gets broken down into the concentration of H⁺ divided by the concentration of H⁺ plus Ka, so that there represents the fraction in the acid form and here the fraction in the conjugate base form or A^- is represented by alpha sub A minus.

Here that would just be equal to Ka divided by H⁺ plus Ka. So these are the 2 equations we can utilize in order to figure out the fraction of either my conjugate acid or my weak acid form, and remember together both of them would equal 1, and if you multiply that by 100, it would be 100. Okay. So if you know one, you know what the other one is because together they represent 100% of all possible molecules and ions present within my solution at any given pH. Now that you see it in terms of a weak monoprotic acid, click on the next video to look at it in terms of a diprotic acid.

Fractional Composition

With diprotic systems, we need to take into account that there are more forms that exist based on pH. With a diprotic system, we have our acidic form where we still possess both of our acidic hydrogens. Once we lose that first acidic hydrogen, we now have the intermediate form. Then, finally, we have our basic form. With these different forms for our diprotic system, we have to take into account more than one \( K_a \) value.

So remember, losing the first acidic hydrogen means we're dealing with \( K_{a1} \), which then gives us this equilibrium expression: HA-=HA×ka1[H+ When we lose that second acidic hydrogen, we have the formation of additional \( H^+ \) as well as the basic form. Here, once we rearrange our first equilibrium expression, we can figure out what our concentration of \( A^{2-} \) is. Now remember, \( HA^- \), our intermediate form equals this entire expression here, which you can then substitute in for this \( HA^- \) here.

So, by substituting in that portion, we get this with only one \( H^+ \) getting multiplied by \( K_{a2} \) plus one \( H^+ \) on the bottom, which is why this is squared here: [A2-=[HA0×ka1×ka2[H+2 From that, we get our formal concentration as being equal to the concentration of our acid form times this large expression here.

The fractions of the three major diprotic forms can be seen as we have our fraction of the acid form, the fraction of our intermediate form, and the fraction of our basic form, each with their own equation. Now, if we take a look here on the left side, we can see how the fraction of each of the forms exists at any given pH. We can see here that this darker line represents the acid form.

This dotted line represents the intermediate form, and then this white, gray line represents the basic form. As the pH is increasing, we're going from left to right, and so the amount of the acid form is decreasing, as well as the intermediate form, and the basic form is increasing. We're going to say, here at this particular pH, we have the intersection between the acid form and the intermediate form. So we'd say here pH equals \( pK_{a1} \) at this intersection. At that intersection, we can also say that the concentration of my acid form is equal to the concentration of my intermediate form.

Then we have this intersection here at a higher pH. At this point, we'd say that pH is equal to \( pK_{a2} \). So at that point, we'd say that the intermediate form concentration is equal to the basic form. So again, we can utilize this chart here to determine what the fraction of two forms are in terms of one another at any given pH. So remember, when it comes to diprotic systems, there are more intermediate forms involved which leads to more complex equations to represent the fractions of each of those forms, and remember your fractional composition is all based on the pH.

The higher the pH, the more the basic forms exist and predominate, and at lower pHs, the acid forms will predominate.

Fractional Composition Calculations 1

Here we're told we have a dibasic compound B that has a pKb1 value of 5 and a pKb2 value of 8. Find the fraction of the acidic form when the pH equals 9.0. Alright. So we're looking for the fraction of the acid form of a dibasic species. That means we're dealing also with the diprotic species looking at it from the opposite direction.

Here that would be the fraction of the acid form equals the concentration of H⁺² divided by the concentration of H⁺² plus the concentration of H⁺ times Ka1 plus Ka1 times Ka2. What we have to do here is determine what our concentration of H⁺ is and determining what my acid dissociation constants k1 and k2 are. We're going to say here that H⁺ equals 10 to the negative pH. H⁺ is equal to 10 to the negative 9.00. So we have that portion figured out so far.

Next, we need to determine what my Ka1 and Ka2 will be. But here we're dealing with pKb1 and pKb2. What we need to remember is that the relationship between your acid dissociation constant, and your base dissociation constant is pKa1 + pKb2 equals 14 and pKa2 + pKb1 equals 14. So Ka1 and Kb2 are connected, and Ka2 and Kb1 are connected.

We're going to say here that pKa1 equals 14 minus pKb2. So that'd be 14 minus 8, which gives us 6. Here, pKa2 equals 14 minus pKb1. So that's 14 minus 5.00, and that's 9. Now that we know pKa1 and pKa2, we know what Ka1 and Ka2 will be.

That's because we're going to say that Ka1 equals 10 to the negative pKa1 and Ka2 equals 10 to the negative pKa2. So plug those values in. So we're going to get 10 to the negative 6.00 and 10 to the negative 9.00. Take those values and plug them in. So that's going to be 10 to the negative 6 and 10 to the negative 9.

10 - 18 ( 2.001 × 10 - 15 )

So at the end, the fraction in my acid form would equal 0.00005. This makes sense because we'd say that at pKa1, our pKa1, we were told, is 6.0. When our pH was equal to our pKa1, we'd have an equal amount of the acid form and the conjugate base form. But as the pH gets higher and higher, we're going to have less and less of the acid form.

The pH is 3 units higher than my pKa1 value of 6. So that means you're going to see a decrease in the amount of my acid form. Being 3 units different is a huge difference in magnitude for concentrations. So it makes sense that my acid form would be such a small value. Remember, dibasic alludes to diprotic.

Here, we're talking about finding the acid form. If they're asking for the intermediate form or the basic form, all we have to do is recall the equations associated with those different forms of diprotic, dibasic species. Utilizing them helps us determine the concentration of any one of those in terms of their fractions. Now that you've seen this example, attempt to do the next one. Where we're talking about the fraction of Tyrosine.

Remember the equations that we've seen on previous pages. Utilize them to find the fraction of that particular amino acid. Once you do, come back and see if your answer matches up with mine.

Fractional Composition Calculations 1

Here it states, what fraction of tyrosine, which has 2 pKa values, exists in all of its forms at pH equals 10.00? Here, because we're dealing with 2 pKa values, that means that Tyrosine exists as a diprotic acid. Tyrosine is one of the amino acids and has 2 acidic Ka values. Remember, for diprotic species, we have the acid form, the intermediate form, and the basic form. Halfway here, we have pH equals pKa1.

At that point, we should have 50% of the acid form and 50% of the basic form or intermediate form actually. And then here halfway, pH equals pKa2. So, we should have 50% of the intermediate form and fifty percent of the basic form. Now in actuality, we won't really have, a flat 50% of both of these at this pH equals pKa1; there will be some weak acid in small minute amounts that we're dealing with.

Realize here that our pKa2 is 8.67 and our pH is above that. What this tells me is it tells me that we've gone beyond this line here, so the predominant form of Tyrosine would be its basic form. What we're going to do now is we're going to use the equations for the fractions of acid form, intermediate form, and basic form to figure out how much we really have of these 3 different forms. For a diprotic species, the fraction in the acidic form equation is the concentration of H⁺ squared divided by the concentration of H⁺ squared plus the concentration of H⁺ times Ka1 plus Ka1 times Ka2 in the denominator. The fraction of the intermediate form equals Ka1 times the concentration of H⁺ divided by the same denominator as the fraction for the acid form.

It's the same exact setup. And then for the fraction of the basic form, it'd be Ka1 times Ka2 divided by the same denominator again. All we have to do is plug in the numbers.

The H⁺ concentration, we get that from the pH because remember, H⁺ concentration equals 10 to the negative pH, which is 10^-10. Then our Ka1, we'd say Ka1 equals 10 to the negative pKa1, which is 10^-2.37. Ka2 equals 10 to the negative pKa2, which is 10^-8.67. When we plug these values in, we'll get the fraction of each one of these 3 different forms for tyrosine. So, here when we plug them in, we'd get 10^-20 divided by when I input each one of these values here into each one of these slots, what I would get at the end would be 9.54669 times 10^-12. Because all three equations have the same exact denominator, all of them would have that same value for the bottom. When I plug this into my calculator, I get 1.05 times 10^-19.

This would be the fraction in the acid form. Look at how small that value is. It makes absolute sense that it would be an incredibly small number because, again, where are we in terms of these 3 different forms? We're around here somewhere because the pH is equal to 10. Our pH is high enough that we're really talking about there being a larger amount of the basic form and the intermediate form.

We've blown past this first line here which represents our pKa of 2.37. Because the pH is so much higher, we're going to have very little of this acid form present within our solution at that pH. And what this really shows us is that no matter what the pH is, we're always going to have some small amount of acid form present within my solution. Now here when we plug in the values, we're going to get 4.2658 times 10^-13 on top divided by the same denominator. So that gives me 0.047 as the fraction of the intermediate form.

And then finally here, we're going to get 9.12011 times 10^-12 divided by the same denominator. That gives me 0.955. What this is showing me is it's showing me that the basic form, of course, is the portion that's the largest in amount, which we expected because the pH is at 10. We've gone beyond this line here, so the major form would be the basic form. Then the next highest form would be my intermediate form with very, very, very small amounts of the acid form here.

If you were to multiply these values by 100, you'd get the percentages of each one, so that would mean that we'd have 95.5 percent in the basic form if we're talking about the percentages of the 3 forms for this diprotic tyrosine compound. Now that we've seen this example, attempt to do the practice question that's left on the bottom of the page. Once you do, come back and see if your answer matches up with mine.

Calculate the fraction of the intermediate for sulfurous acid, H ₂SO₃, at pH = 11.00? pK_a1 = 1.80, pK_a2 = 7.19. 

• A. 6.31x10^-11
• B. 9.77x10^-14
• C. 0.000155
• D. 0.00977
• E. 1.00

Answer: (C)

Fractional Concentrations

So recall when it comes to monoprotic systems that the fraction of the acid molecules and the conjugate base molecules were represented by αa and αa-. Here when we're talking about the fraction of acid molecules, it would equal the h+ concentration divided by [H⁺][H⁺] + K_a and the fraction of the base concentration was equal to Ka[H⁺] + K_a. As we look at this chart here on the left, as our pH increased going from left to right, the amount of the acid molecules would decrease as the amount of conjugate base molecules would increase. Here at this intersection is where both the concentration of the HA molecules and A- molecules would be equal to one another. Also remember that that happens when pH is equal to the pKa.

Now when we try to determine what our concentrations are for these fractions of these molecules, we can restructure our equations to now represent the concentration of my acid is equal to the fraction of the acid molecules times the formal concentration of the acid molecules. Using this expression now inputs the formal concentration of my weak acid solution and if we're looking for the overall concentration of my conjugate base, it's equal to the fraction of my conjugate base molecules times the formal concentration of my solution, acid solution, so that'd be Ka1[H⁺] + K_a times formal concentration of acid again divided by [H⁺] + K_a. So remember the fraction can be determined by comparing the pH of the solution to the pKa value to see which species is greater in amount. We can utilize these equations themselves to find either the exact fraction of the weak acid or the exact fraction of the conjugate base. From there, we can attach formal concentration to find the new concentrations of the weak acid and the conjugate base.

Click on to the next video and see what we do when dealing with diprotic systems.

Fractional Concentrations

So as we stated before in the past when it comes to diprotic systems, we have 3 possible forms that our compound can take. We have the acidic form, we have the intermediate form after it's lost its first acidic hydrogen, and then we have the basic form. Here with these 3 different forms, we have the introduction of Ka1 and Ka2. When removing the first acidic hydrogen, we produce your hydronium ion plus some intermediate. Ka1 would be involved which would give us this expression of products over reactants.

By rearranging it, we'd be able to isolate the concentration of our intermediate as the concentration of your acid form times Ka1 divided by H+. Here from the intermediate form, we could lose our second acidic hydrogen to give us more hydronium ion created, plus now the basic form of our compound. By substituting in this expression for our intermediate, we can get the concentration of our basic form as being HA2⁢Ka1⁢Ka2(H+)2. Now our formal concentration comes about through the use of our mass balance. Here we'd have our formal concentration equal to the concentration of my acid form times the expression within here.

Now remember the full concentrations of the 3 major diprotic forms can be restructured as follows. We've seen the equations related to the fractions of the acid form, the intermediate form, and the basic form. With the introduction of our formal concentration here, we'll be able to calculate the new concentrations of each one of these compounds. By doing that, we just have the insertion of the formal concentration into each one of these compounds, and remember all three forms, although different, possess the same denominator on the bottom. Also recall that when we're taking a look at the fraction versus pH chart, that this line here represents the amount or concentration of my acid form.

This here represents the concentration of my intermediate form, and then this one here represents the concentration of my basic form. At this intersection, we have pH=pKa1, because the concentration of H2A would be equal to the concentration of HA-. And here at our second intersection, we'd have pH=pKa2 because the intermediate form would be equal to the basic form. This just shows us as the pH is increasing going from left to right, the proportion of the conjugate or the basic form is increasing, and remember even at high enough pHs, we're still going to have some amount of our acidic form remaining. It'll just be a very, very, very small amount the higher the pH gets, and remember all three forms exist in some proportion the higher the pH gets.

So we've seen these equations in terms of the now we're introducing our formal concentrations to find our new conditional concentration of our acid form, our intermediate form, and the basic form.

Fractional Concentrations Calculations 1

Here it states a dibasic compound B has a pK_b1 value of 4.00 and a pK_b2 value of 6.00. We're asked to find the concentration of the intermediate form when the formal or total concentration of the diprotic acid equals 0.150 molar and the pH equals 8.00. Alright. So, here we need to determine the concentration of our intermediate. So we're going to say the concentration of our intermediate form equals the fraction of the intermediate form times the formal concentration of the diprotic acid.

When we expand this out, it gives me K_a1 times [H⁺] times the formal concentration of my diprotic acid divided by [H⁺]² + [H⁺] times K_a1 + K_a1 times K_a2. Alright. So we know what [H⁺] is equal to because [H⁺] concentration equals 10 to the negative pH. And in this case, we're dealing with a pH of 8.00. Next, we need to figure out what K_a1 and K_a2 will be.

We're gonna say that pK_a1 + pK_b2 and that pK_a2 + pK_b1 equals 14. So pK_a1 here equals 14 minus pK_b2, which is 6. So this comes out to being 8. Because we now know what pK_a1 is, we know what K_a1 is. K_a1 equals 10 to the negative pK_a1.

So that's 10 to the negative 8.00. pK_a2 will equal 14 minus pK_b1, so it equals 10. So K_a2 equals 10 to the negative pK_a2, so it's 10 to the negative 10. We now have all the data that we need, so we're just going to plug them into the formula above. So K_a1 is 10 to the negative 8 times 10 to the negative 8 times the formal concentration of 0.150 molar, dividing now by 10 to the negative 8.00 squared plus 10 to the negative 8.00 times K_a1 plus the 2 K_a's multiplying each other. So 10 to the negative 8.00 times 10 to the negative 10. When we figure out what's on the top and on the bottom, we're going to get 1.5 times 10 to the negative 17 divided by 2.01 times 10 to the negative 16.

So at a pH of 8.0, the amount of it that exists in the intermediate form, concentration-wise, equals 0.075 molar. This would be our amount of the intermediate form. Remember, when we're dealing with our diprotic or dibasic compound, we have three states to consider: the acid form, the intermediate form, and the basic form. Here, pH equals pK_a2 which is 6.

But our pH is above 6; our pH is 8, which means that a majority of this formal concentration of my diprotic acid would actually exist in the basic form. A majority of it exists in that form, and a portion of it exists in the intermediate form and the acidic form. What we just found was the amount that exists in the intermediate form, starting from the initial formal concentration of 0.150 molar. As the pH would increase, this number would decrease even more because the higher the pH gets, the more the basic form will become the predominant concentration involved in my solution. If we were to find the concentration of my acidic form, it would be even less than this value here.

Now that you've seen this example in terms of determining the concentration, look to see if you can answer example 2 that's given down below. Once you do, come back and see if your answer matches up with mine.

Fractional Concentrations Calculations 1

Here, it tells us to calculate the concentration of the acidic form for 0.230 molar tartaric acid when the pH is 6.00. We're given 2 pKa values. pKa1 equals 3 and pKa2 equals 4.34. Since they're telling us to calculate the concentration, we'll assume that this number here must be the full concentration of my diprotic acid. We're going to say here that the concentration of my diprotic acid equals the fraction of my diprotic acid times the formal concentration of my diprotic acid.

This translates to h⁺ squared times the formal concentration of my acid divided by h+ squared plus h+Ka1 plus ka2 ka1. Remember here that h⁺ is equal to 10 to the negative pH. So, it equals 10 to the negative 6. K၁₁ equals 10 to the negative pKa၁₁, so it equals 10 to the negative 3. And K၁₂ equals 10 to the negative pKa၁₂, so it equals 10 to the negative 4.34.

All we do now is we insert our values. So, h⁺ squared, so that's 10 to the negative 6, squared times the formal concentration divided by h⁺ squared plus h⁺ times Ka၁₁ which is 10 to the negative 3 plus 10 to the negative 3 times 10 to the negative 4.34. When we figure out the values for top and bottom, we get 2.30 times 10 to the negative 13 divided by 4.67098 times 10 to the negative 8. So, that gives me a fraction of or concentration of 4.92 times 10 to the negative 6 molar. So, our concentration here is incredibly small because with our diprotic acid, we'd have the acidic form, we'd have the intermediate form, and we'd have the basic form.

Here, pH would equal pKa1, and here, pH would equal pKa2. Here, at pKa2, it is equal to 4.34. That's when we would have approximately 50% of the basic form and 50% of the intermediate form. Our pH is even higher than that. It's at 6, so we'd have a majority of my solution composed of the basic form.

The intermediate form will form the 2nd most amount, and because the acidic form is at 50% around a pH of 3, we'd expect very little of it when the pH is all the way up to 6. What this shows us is that although the pH is incredibly high, we never totally get rid of all of our acidic form. There'll still be small amounts of it present within our solution. They'll just be incredibly small amounts. That explains why our concentration here is times 10 to the negative 6 molar.

The other 2 would be much greater in amount because the pH again is at 6 and at pH 4.34, we'd have 50% of both of these. Since the pH is greater than 6, the majority of our solution exists in the basic form. Remember the formulas that are associated with each one of the forms for a diprotic system and remember what our pH is telling us. It's giving us the fraction distribution between the 3 different forms for my diprotic species.

Here’s what students ask on this topic:

What is the relationship between pH and pKa in determining the fractional composition of weak acids and their conjugate bases?

The relationship between pH and pKa is crucial in determining the fractional composition of weak acids and their conjugate bases. When pH equals pKa, the concentrations of the weak acid (HA) and its conjugate base (A⁻) are equal. This is derived from the Henderson-Hasselbalch equation: ${\mathrm{pH}}_{}={\mathrm{pKa}}_{}+\log \left(\frac{{\mathrm{[A⁻]}}_{}}{{\mathrm{[HA]}}_{}}\right)$. At this point, the log term becomes zero, making pH equal to pKa. As pH increases above pKa, the fraction of the conjugate base increases, and the fraction of the weak acid decreases.

How do you calculate the fractional composition of a diprotic acid system?

In a diprotic acid system, you need to consider both dissociation constants, Ka1 and Ka2. The fractional compositions of the acidic form (H₂A), intermediate form (HA⁻), and basic form (A²⁻) are calculated using specific equations. For the acidic form: ${\alpha }_{=}$. For the intermediate form: ${\alpha }_{=}$. For the basic form: ${\alpha }_{=}$. These equations help determine the fraction of each form at any given pH.

What is the significance of the Henderson-Hasselbalch equation in fractional composition calculations?

The Henderson-Hasselbalch equation is significant in fractional composition calculations because it relates pH, pKa, and the ratio of the concentrations of the conjugate base and weak acid. The equation is: ${\mathrm{pH}}_{}={\mathrm{pKa}}_{}+\log \left(\frac{{\mathrm{[A⁻]}}_{}}{{\mathrm{[HA]}}_{}}\right)$. This equation allows you to determine the pH of a solution if the concentrations of the acid and its conjugate base are known, or vice versa. It is particularly useful for buffer solutions, where the pH is maintained within a narrow range. When pH equals pKa, the concentrations of the acid and conjugate base are equal, which is a key point in understanding fractional compositions.

How do you determine the concentration of a weak acid and its conjugate base at a given pH?

To determine the concentration of a weak acid (HA) and its conjugate base (A⁻) at a given pH, you can use the fractional composition equations. For the weak acid: ${\mathrm{[HA]}}_{}=\frac{{\mathrm{[H⁺]}}_{}{F}_{}}{{\mathrm{[H⁺]}}_{}+{\mathrm{Ka}}_{}}$. For the conjugate base: ${\mathrm{[A⁻]}}_{}=\frac{{\mathrm{Ka}}_{}{F}_{}}{{\mathrm{[H⁺]}}_{}+{\mathrm{Ka}}_{}}$. Here, F is the formal concentration of the solution. These equations allow you to calculate the exact concentrations of the weak acid and its conjugate base at any given pH, helping you understand the solution's composition.

What are the key differences between monoprotic and diprotic acid systems in terms of fractional compositions?

The key differences between monoprotic and diprotic acid systems in terms of fractional compositions lie in the number of dissociation steps and the complexity of the equations. Monoprotic acids have one dissociation constant (Ka) and two forms: the acid (HA) and its conjugate base (A⁻). The fractional compositions are simpler and can be calculated using: ${\alpha }_{=}$ for the acid and ${\alpha }_{=}$ for the conjugate base. Diprotic acids have two dissociation constants (Ka1 and Ka2) and three forms: the acid (H₂A), the intermediate (HA⁻), and the base (A²⁻). The fractional compositions are more complex and require additional equations to account for the intermediate form. This makes diprotic systems more intricate to analyze compared to monoprotic systems.