Ionic Strength of Soluble Salts - Online Tutor, Practice Problems & Exam Prep

Silver bromide, an ionic solid, dissociates into ions in solution, establishing an equilibrium characterized by a low solubility product constant (Ksp). The common ion effect reduces solubility when additional ions are present, as per Le Chatelier's principle. Conversely, noncommon ions can increase solubility by enhancing ionic strength, which is calculated using the equation: μ2=(1/2)(Σ([C_i]⁢z_i)). Understanding these concepts is crucial for predicting solubility behavior in various solutions.

Ionic Strength

So let's take a look at the association of silver bromide. Silver bromide is an ionic solid. When we throw it into solution, it's going to basically dissociate into its ions. But in the process, an equilibrium is established. That just means that very little of these ions are produced during this process. Now, connected to the solubility of any ionic solid is our \( K_{sp} \) value, our solubility product constant. Remember, we've talked about this in the past. When it comes to the \( K_{sp} \) values for ionic solids, they're typically much less than 1 because these ionic solids have minimal solubility within a solvent. Now, let's say to this pure solution of water that has this ionic compound dissolving, we will produce \( x \) amount of these ions. Now let's say we took that same silver bromide and instead it was common. They both have bromide ion involved. Remember, when it comes to \( K_{sp} \) of ionic solids, we have initial change and equilibrium involved. Here, the common ion we have is bromide. Initially, this would be 0.10 molar. This has no common ion right now, so we'd say that's 0. Remember, we call this the common ion effect where we'd have an initial amount of 1 for both ions depending on the solution present. Remember, this type of situation helps to decrease the overall solubility of my ionic compound. Remember, think of Le Chatelier's principle. We're adding more of this ion here. According to Le Chatelier's principle, when I add product, I have to move in reverse to get rid of it. This will cause my reaction to favor the reverse direction towards the solid once again. Remember, solubility is how much of this ionic solid can we get to dissolve. If it's moving backwards, it's not dissolving. It's crystallizing. Same thing would happen if I had 0.25 molar of silver acetate. Here, the common ion is silver. In that solution, we'd have an initial amount of 0 point 25 molar of silver, again falling into common ion effect which again would lower my overall solubility of my ionic compound. With these common ions, we have the overall solubility decreasing as a result of the common ion effect. But let's say we didn't put it in a solution that had a common ion. Let's say instead, we put it in 0.01 Molar Sodium Perchlorate. Now, here sodium is the positive ion and chlorate is the negative ion. Neither one of those ions matches up with the ions associated with my \( K_{sp} \) value. Here, there is no common ion effect involved. What would happen is that this positive ion here would surround this bromide ion and they basically, it would in a way decrease the amount of free floating bromide ions within the solution. The same way here, this perchloride ion would surround this silver ion. As a result, I need to make more of these ions because they're being surrounded by these non common ions here. So to make more of the ions, the ionic solid has to move forward by Le Chatelier's principle to remake more of those free floating product ions. So here, if we put in ions that are not common to our equation, the overall solubility of ionic compound will increase as a result of ionic strength. So ionic strength is just the measurement of all the ions in the aqueous solution and ionic strength itself has an equation. Here, we're going to say ionic strength represents the interactions between ions in water and ions in the solution. Ionic strength, which is \( \mu \), equals half the summation of the concentration of the ions times their charges squared. If you have 2 ions, it would be the concentration of the first ion times its charge squared (represented in MathML as \( \text{concentration} \times q^2 \)) plus the concentration of the second ion times its charge squared. If we add additional ions, we just keep going until we've found the ionic strength. Remember, the common ion effect helps to decrease the overall solubility of an ionic compound, but non-common ions help to increase solubility because of ionic strength. They surround those free floating ions, and so your ionic solid adjusts by moving in the forward direction to create even more of those ions. Now that we know the basic understanding of ionic strength, take a look at the example that's left below. Once you're done, click over to the next video and see how I approach in calculating the ionic strength of that particular compound.

Ionic Strength

Here we have to calculate the ionic strength of the following ionic compound. This compound is copper. Now, how do I know that copper has a plus 2 charge? Well, you need to remember your polyatomic ion. SO₃ here is the sulfide ion. That's the formula of the sulfide ion. If the overall compound is neutral and it's a one-to-one relationship, 1 copper to 1 sulfide, then if this is 2 minus, copper has to be 2 plus. The charges cancel out and I have a neutral ionic compound at the end. So right now, we have a one-to-one relationship of the ions. If the concentration for the entire ionic compound is 0.010 molar, then each ion produced will have that same concentration. Here, this is 0.010 molar and this is 0.010 molar. Now, we're going to say that our ionic strength equals half the sum of the product of the concentration of the first ion, which is 0.010, times its charge squared, so +2 squared, plus the concentration of the second ion times its charge squared. When we do all that, we'll get an answer of 0.04 for the ionic strength of 0.010 molar copper(II) sulfide. So just follow the steps that we did. If there were multiple ions, so let's say that there were 2 coppers for some reason, then we'd have to multiply the concentration by 2 because we have to take into account all the copper ions with the overall concentrations that they have. But here, since it's a one to one, the initial concentration I have for the compound matches the concentrations for the ions provided. Later on, we'll continue more questions that deal with determining the ionic strength of an overall ionic compound. Again, make sure you remember your polyatomic ions and how to correctly classify the relationships between the ions and their charges involved.

Ionic Strength Calculations

Just as we stated in the previous videos, when figuring out the ionic strength for any ionic compound, it's best to break it up into its ions to see how many ions we have of each one so we can figure out their concentrations. Here we have aluminum carbonate. Aluminum carbonate breaks up into 2 aluminum ions plus 3 carbonate ions. The concentration of the entire ionic compound is this. But the number of ions of each type determines their overall concentration. Here we have 2 aluminums, so it'd be 2 times the original concentration. The overall concentration of aluminum ions would be 0.060 molar. Here, we have 3 carbonates, so that's 3 times the original concentration. The overall concentration of carbonate is 0.090 molar. Now, we use the formula to determine ionic strength. So, ionic strength equals half. Remember, it's the summation of the concentration of the ion times its charge squared plus the concentration of the next ion times its charge squared. So that'll be:

Ionic Strength = 1 2 ( [ Al 3+ ] ∙ 3 2 + [ CO 3 − 2 ] ∙ 2 2 )

Here, if you work out what's in the brackets correctly, you'll get half of 0.90. That would mean that my ionic strength here would be 0.45. Now that you've seen this example, try the next one. In this one, we're asked to figure out the total ionic strength of this solution. Here we have 2 different ionic compounds mixing together. What effect would that have on the concentration of each ion present to determine the final ionic strength? Try it out for yourself and then come back and see how I approach this same example to question.

Ionic Strength Calculations

Here it asks, what is the ionic strength of a solution that is 0.1 molar sodium phosphate and 0.05 molar sodium hydrogen phosphate. All right. We're going to break these compounds up into their ions. Here, sodium phosphate breaks up into 3 sodium ions plus 1 phosphate ion. Sodium hydrogen phosphate breaks up into 2 sodium ions plus 1 hydrogen phosphate ion. Remember, the number of each particular ion has an effect on the overall concentration that it has. Here, the overall molarity of the entire sodium phosphate molecule or compound is 0.1 molar, but there are 3 sodiums here, so that's 3 times 0.1 molar. Here this would just be 0.1 molar since there's only 1. Here, this is 0.05 molar for this, so this is 2 times 0.05 molar since there are 2 sodiums. And here, this would just be 0.05 molar. For ionic strength, remember, ionic strength equals half, and it's the summation of concentration times its charge squared.

Now, I know that both of these are sodium, so technically, if you want, you could combine them together. But I'm keeping them separate because they have different concentrations.

So the formulation for ionic strength can be given by: I = 1 2 ∑ c i z 2 where ci and z are concentration and charge of each ion, respectively. For this solution: I = 0.5 ( 0.3 × 1 +2 + 0.1 × 3 -2 + 0.1 × 1 +2 + 0.05 × 2 +2 ) That's going to give a final ionic strength of 0.75.

Now that you've seen these two examples, attempt this practice question that's left here on the bottom. Remember, the first thing you should do is to split this compound into its ions. From there, determine how many you have of each particular ion. Then from that, you can determine their concentrations to help you determine the final ionic strength for the entire ionic compound. So guys, attempt it on your own. Once you've done that, come back and see how I approach the same exact practice question.

Calculate the ionic strength for the following ionic compound.

0.04 M SnO₂

• A. 0.40
• B. 0.48
• C. 0.12
• D. 0.96
• E. 0.24

Here’s what students ask on this topic:

What is the common ion effect and how does it influence the solubility of ionic compounds?

The common ion effect refers to the decrease in solubility of an ionic compound when a solution already contains one of the ions present in the compound. According to Le Chatelier's principle, adding a common ion shifts the equilibrium towards the solid form, reducing the amount of the compound that can dissolve. For example, adding bromide ions to a solution of silver bromide will decrease its solubility because the equilibrium will shift to form more solid silver bromide.

How is ionic strength calculated in a solution?

Ionic strength (μ) is a measure of the total concentration of ions in a solution and their interactions. It is calculated using the equation:

$\mu =\frac{1}{2}\sum (C_i\cdot {z_i}^2)$

where C_i is the concentration of the i-th ion and z_i is its charge. The sum is taken over all ions in the solution. This equation accounts for both the concentration and the charge of each ion, reflecting their contribution to the overall ionic strength.

What is the solubility product constant (Ksp) and how is it used?

The solubility product constant (Ksp) is a measure of the solubility of an ionic compound in a solvent. It is the product of the molar concentrations of the constituent ions, each raised to the power of their stoichiometric coefficients. For a compound AB that dissociates into A⁺ and B^-, the Ksp is given by:

$K_{\mathrm{sp}}=\left[A\right]\left[B\right]$

Ksp values are typically much less than 1, indicating low solubility. It is used to predict whether a precipitate will form in a solution and to calculate the solubility of the compound under different conditions.

How do noncommon ions affect the solubility of ionic compounds?

Noncommon ions can increase the solubility of ionic compounds by enhancing the ionic strength of the solution. These ions do not participate in the equilibrium of the dissolving ionic compound but surround the free-floating ions, reducing their activity. According to Le Chatelier's principle, the system compensates by dissolving more of the ionic solid to produce more free ions. This results in an increase in the overall solubility of the compound.

What role does Le Chatelier's principle play in the solubility of ionic compounds?

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system responds to counteract the change and restore a new equilibrium. In the context of solubility, adding a common ion shifts the equilibrium towards the solid form, reducing solubility. Conversely, adding noncommon ions increases ionic strength, causing the equilibrium to shift towards dissolving more of the solid to produce more free ions, thus increasing solubility.