Ionic Strength of Soluble Salts: Videos & Practice Problems

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

Silver bromide (AgBr) is an ionic solid that dissociates into its constituent ions when placed in a solution, establishing an equilibrium. The solubility product constant, K_sp, is crucial in understanding the solubility of ionic compounds like silver bromide, which typically has a K_sp value less than 1, indicating low solubility in water.

When silver bromide is dissolved in a solution containing a common ion, such as sodium bromide (NaBr), the solubility of silver bromide decreases due to the common ion effect. In this case, the initial concentration of bromide ions is 0.10 M, which shifts the equilibrium according to Le Chatelier's principle. The addition of more bromide ions causes the reaction to favor the formation of solid silver bromide, thus reducing the amount of dissolved ions.

Similarly, if silver bromide is placed in a solution of silver acetate (AgC₂H₃O₂), the presence of silver ions (Ag⁺) also decreases the solubility of silver bromide due to the common ion effect.

Conversely, when silver bromide is introduced into a solution like 0.01 M sodium perchlorate (NaClO₄), which does not contain any common ions, the solubility of silver bromide increases. This occurs because the non-common ions surround the bromide and silver ions, effectively reducing their concentration in the solution. To restore equilibrium, the dissolution of silver bromide is favored, leading to an increase in the number of free ions in solution.

The concept of ionic strength is essential in this context, as it quantifies the interactions between ions in a solution. Ionic strength (I) can be calculated using the formula:

I = 1/2 Σ(c_i * z_i²)

where c_i is the concentration of each ion and z_i is the charge of each ion. This measurement helps to understand how the presence of various ions affects the solubility of ionic compounds.

In summary, the common ion effect decreases the solubility of ionic compounds, while the presence of non-common ions can enhance solubility by increasing ionic strength, thereby promoting the dissolution of the ionic solid.

Ionic Strength

To calculate the ionic strength of copper(II) sulfite, we first need to understand how the compound dissociates in solution. Copper(II) sulfite breaks down into copper(II) ions (Cu²⁺) and sulfite ions (SO₃^2-). The charge of copper is determined by the charge of the sulfite ion; since the overall compound is neutral and the sulfite ion has a charge of 2-, the copper must have a charge of 2+ to balance it out.

Given that the concentration of copper(II) sulfite is 0.010 M, each ion produced will also have a concentration of 0.010 M. Therefore, we can express the ionic strength (I) using the formula:

I = 0.5 * (C₁ * z₁² + C₂ * z₂²)

In this case, C₁ is the concentration of copper(II) ions (0.010 M) and z₁ is the charge of the copper ion (+2), while C₂ is the concentration of sulfite ions (0.010 M) and z₂ is the charge of the sulfite ion (-2). Plugging in these values, we have:

I = 0.5 * (0.010 * (2)² + 0.010 * (2)²)

Calculating this gives:

I = 0.5 * (0.010 * 4 + 0.010 * 4) = 0.5 * (0.04 + 0.04) = 0.5 * 0.08 = 0.04

Thus, the ionic strength of a 0.010 M solution of copper(II) sulfite is 0.04. If there were multiple ions of a particular type, such as two copper ions, we would adjust the concentration accordingly to account for the total contribution to ionic strength. Understanding the relationships between ions and their charges is crucial for accurately calculating ionic strength in various ionic compounds.

Ionic Strength Calculations

To calculate the ionic strength of an ionic compound, it is essential to first dissociate the compound into its constituent ions. For instance, aluminum carbonate (Al₂(CO₃)₃) dissociates into 2 aluminum ions (Al³⁺) and 3 carbonate ions (CO₃^2-). The overall concentration of the compound influences the concentration of each ion type. If the concentration of aluminum carbonate is given as a certain value, the concentration of aluminum ions would be calculated as twice that value, while the concentration of carbonate ions would be three times the original concentration.

For example, if the concentration of aluminum carbonate is 0.030 M, the concentration of aluminum ions would be:

Concentration of Al³⁺ = 2 × 0.030 M = 0.060 M

And for carbonate ions:

Concentration of CO₃^2- = 3 × 0.030 M = 0.090 M

Next, to find the ionic strength (I) of the solution, we use the formula:

I = 1/2 Σ(c_i z_i²)

Where c_i is the concentration of each ion and z_i is the charge of each ion. For aluminum ions, the charge is +3, and for carbonate ions, the charge is -2. Plugging in the values:

I = 1/2 [(0.060 M)(3²) + (0.090 M)(-2²)]

This simplifies to:

I = 1/2 [(0.060 M)(9) + (0.090 M)(4)]

I = 1/2 [0.540 + 0.360]

I = 1/2 [0.900] = 0.45

Thus, the ionic strength of the solution is 0.45 M. Understanding how to calculate ionic strength is crucial, especially when dealing with solutions containing multiple ionic compounds, as the presence of different ions can affect the overall ionic strength significantly. In such cases, it is important to consider the concentration and charge of each ion from all compounds present to determine the final ionic strength accurately.

Ionic Strength Calculations

To calculate the ionic strength of a solution containing sodium phosphate and sodium hydrogen phosphate, we first need to dissociate these compounds into their respective ions. Sodium phosphate (Na₃PO₄) dissociates into three sodium ions (Na⁺) and one phosphate ion (PO₄^3-). Sodium hydrogen phosphate (Na₂HPO₄) dissociates into two sodium ions (Na⁺) and one hydrogen phosphate ion (HPO₄^2-).

Next, we determine the concentration of each ion based on the molarity of the compounds:

• For sodium phosphate (0.1 M):
  • 3 Na⁺: 3 × 0.1 M = 0.3 M
  • 1 PO₄^3-: 0.1 M
• For sodium hydrogen phosphate (0.05 M):
  • 2 Na⁺: 2 × 0.05 M = 0.1 M
  • 1 HPO₄^2-: 0.05 M

Now, we can combine the concentrations of the ions:

• Total Na⁺: 0.3 M (from sodium phosphate) + 0.1 M (from sodium hydrogen phosphate) = 0.4 M
• PO₄^3-: 0.1 M
• HPO₄^2-: 0.05 M

The formula for ionic strength (I) is given by:

I = \frac{1}{2} \sum c_{i} z_{i}^{2}

Where:

• c_i = concentration of each ion
• z_i = charge of each ion

Substituting the values into the ionic strength formula:

I = \frac{1}{2} \left( (0.4 \, \text{M} \cdot 1^{2}) + (0.1 \, \text{M} \cdot (-3)^{2}) + (0.1 \, \text{M} \cdot 2^{2}) + (0.05 \, \text{M} \cdot (-2)^{2}) \right)

Calculating each term:

• Na⁺: 0.4 M × 1² = 0.4
• PO₄^3-: 0.1 M × 3² = 0.9
• HPO₄^2-: 0.1 M × 2² = 0.4
• Na⁺: 0.1 M × 1² = 0.1

Now, summing these values gives:

I = \frac{1}{2} (0.4 + 0.9 + 0.4 + 0.1) = \frac{1}{2} (1.8) = 0.9

Thus, the ionic strength of the solution is 0.9 M. Understanding how to break down compounds into their ions and calculate their contributions to ionic strength is crucial for various applications in chemistry, particularly in solutions and electrochemistry.

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

Answer: (B)

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.