Question

A solution contains 0.018 mole each of \(\mathrm{I}^{-}, \mathrm{Br}^{-},\) and \(\mathrm{Cl}^{-}.\) When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24 M\) \(\mathrm{AgNO}_{3}\) what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\mathrm{AgI} : K_{\mathrm{sp}}=1.5 \times 10^{-16}$$ $$\operatorname{AgBr} ; K_{\mathrm{sp}}=5,0 \times 10^{-13}$$ $$\mathrm{AgCl} : K_{\mathrm{sp}}=1.6 \times 10^{-10}$$

Short answer

In conclusion, 0.85992 g of AgCl(s) will precipitate out, and the final concentration of Ag+ ions is 0.03 M.

Step-by-step solution

Determine the initial concentrations of ions

First, we need to determine the initial concentrations of I-, Br-, and Cl- ions in the solution. Since there are 0.018 moles of each ion in the solution, we can calculate the initial concentration of these ions by dividing the number of moles by the total volume of the solution (200 mL). \[ [\mathrm{I}^{-}]_0 = [\mathrm{Br}^{-}]_0 = [\mathrm{Cl}^{-}]_0 = \frac{0.018 \ \text{moles}}{0.200 \ \text{L}} = 0.09 M \]

Determine the ion product of each silver salts

Then, we need to find the initial ion product of each silver salt using the concentration of Ag+ ions and the concentration of each anion. Ion product (IP) for each silver salt is given by: \[{\text{IP}}_{\text{AgI}} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}] \] \[{\text{IP}}_{\text{AgBr}} = [\mathrm{Ag}^{+}][\mathrm{Br}^{-}] \] \[{\text{IP}}_{\text{AgCl}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \] Since initially, the Ag+ ions are not in the solution, the concentration of Ag+ ions should be considered as 0.24 M (the concentration of AgNO3 in the 200 mL solution). So the ion products will be: \[{\text{IP}}_{\text{AgI}} = 0.24 \times 0.09 \] \[{\text{IP}}_{\text{AgBr}} = 0.24 \times 0.09 \] \[{\text{IP}}_{\text{AgCl}} = 0.24 \times 0.09 \] As we can see, the ion products for all the salts will have the same value.

Identify the first salt to precipitate

Now we want to compare the calculated ion products with their Ksp values, to see which salt will precipitate first. Since IP is equal for all three salts, compare the IP with the smallest Ksp first. The smallest Ksp belongs to AgI (Ksp = 1.5 × 10^-16). \[ {\text{IP}}_{\text{AgI}} = 0.24 \times 0.09 = 0.0216 \] Since IP_AgI > Ksp_AgI, AgI(s) will precipitate first.

Calculate the moles of remaining Ag+ ions

Since all of the 0.018 moles of I- ions will react with Ag+ ions and form AgI precipitate: \[ \text{moles of remaining Ag}^{+} = 0.018 \ \text{moles} - 0.018 \ \text{moles} = 0.006 \ \text{moles} \]

Calculate the remaining concentration of Ag+ ions

Next, we need to find the remaining concentration of the Ag+ ions in the solution. Divide the moles of remaining Ag+ ions by the total volume of the solution to get the concentration. \[ [\mathrm{Ag}^{+}]_\text{final} = \frac{0.006 \ \text{moles}}{0.200 \ \text{L}} = 0.03 M \]

Calculate the mass of AgCl that precipitates out

Since AgI has already started to precipitate, now we want to check if AgBr and AgCl will also precipitate. We know that the remaining concentration of Ag+ ions is lower than the initial concentration. So, check the ion product for the AgCl since it has the highest Ksp among the three salts. \[{\text{IP}}_{\text{AgCl, new}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] = 0.03 \times 0.09 = 0.0027 \] Since \({\text{IP}}_{\text{AgCl, new}} > {\text{Ksp}}_{\text{AgCl}}\), AgCl will also precipitate. Now, we want to calculate how many moles of AgCl(s) are formed. As all the remaining 0.006 moles of Ag+ ions will react with Cl- ions to form AgCl(s), 0.006 moles of AgCl(s) will form. Now, we can calculate the mass of AgCl(s) by multiplying the moles by the molar mass of AgCl. \[ \text{Mass of AgCl} = \text{moles of AgCl} \times \text{Molar mass of AgCl} \] \[ \text{Mass of AgCl} = 0.006 \ \text{moles} \times 143.32 \frac{\text{g}}{\text{moles}} = 0.85992 \ \text{g} \] In conclusion, 0.85992 g of AgCl(s) will precipitate out, and the final concentration of Ag+ ions is 0.03 M.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, represented as \(K_{sp}\), is a measure of the solubility of a compound under specific conditions. It's particularly relevant for sparingly soluble salts. When a solid is in equilibrium with its ions in the solution, \(K_{sp}\) reflects this balance. For example, in the exercise above, we consider salts like AgI, AgBr, and AgCl. Each has a different \(K_{sp}\) value, which tells us how easily that salt can dissolve in water. To determine if a salt will precipitate, compare its ion product (IP) to its \(K_{sp}\). If \(IP > K_{sp}\), the solution is supersaturated, and the salt will precipitate. Conversely, if \(IP < K_{sp}\), it's unsaturated, and no precipitation occurs. In our example, \(IP\) calculations help determine which silver halide precipitates first when mixing with AgNO3. Understanding \(K_{sp}\) allows us to predict whether a compound will stay dissolved or form a solid. Recognizing this balance is essential when working with reactions involving limited solubility.

Ion Concentration

Ion concentration refers to the amount of a particular ion present in a solution. It is usually expressed in molarity (M), which is moles per liter. In our exercise, we start with the molarities of \([\mathrm{I}^-]\), \([\mathrm{Br}^-]\), and \([\mathrm{Cl}^-]\), each at 0.09 M. To find ion concentrations when additional reactants are introduced, such as AgNO3, you'll first need to determine final concentrations. When mixing solutions, multiply the molarity of each ion by the overall volume to get the initial mole count. Then divide this by the new total volume if the volume changes. However, in this problem, we assume no volume change, simplifying calculations. The ion product, a key concept in determining whether a precipitate forms, involves the multiplication of concentrations. It shows how concentrated the reactants are in relation to their equilibrium state. Knowing these concentrations allows for comparisons to \(K_{sp}\) values, revealing if a precipitate like AgCl will form.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually measured in grams per mole (g/mol). It is crucial for converting between the mass of a compound and the number of moles during chemical calculations. For salts like AgCl, we use molar mass to determine the mass of the precipitate forming. Here’s a simple way to calculate it:

• Find the atomic masses of each element in the compound from the periodic table.
• Add up these atomic masses according to the formula of the compound.

For AgCl:

• Silver (Ag) has an atomic mass of approximately 107.87 g/mol.
• Chlorine (Cl) has an atomic mass of approximately 35.45 g/mol.
• Add them together to get the molar mass of AgCl: 143.32 g/mol.

In the exercise, we found 0.006 moles of AgCl formed, which when multiplied by 143.32 g/mol gives the mass of precipitated AgCl as 0.85992 g. Molar mass provides a bridge from the conceptual (moles) to the tangible (grams), enabling precise measurements in laboratory settings. Understanding molar mass is fundamental for calculating the amounts of substances in reactions.