a quantity of 0.5 g of crystalline NaCl is added to one liter of soultion containing 10e-3 mol of dissolved Ag+ demonestrate that AgCl precipitate from the solution and calculate the wight og AgCl in grams that form per liter. Assume that all activities coefficients are equal to one.
To demonstrate that AgCl precipitates from the solution, we need to determine if the product of the ion concentrations of Ag+ and Cl- exceeds the solubility product constant (Ksp) of AgCl. If the product exceeds the Ksp, AgCl precipitates.
Let's start by calculating the concentration of Cl- ions in the solution. Given that the solution contains 10^-3 mol of dissolved Ag+, it means that there is also 10^-3 mol of Cl- ions since they combine in a 1:1 ratio with Ag+ to form AgCl.
Next, we need to calculate the concentration of Ag+ ions that will react with the Cl- ions to form AgCl. Since 0.5 g of NaCl is added to the solution, we need to convert it to moles. The molar mass of NaCl is 58.44 g/mol, so:
0.5 g / 58.44 g/mol = 0.00855 mol
Since NaCl dissociates into one Na+ ion and one Cl- ion when dissolved in water, the concentration of Ag+ ions is also 0.00855 mol.
Now, let's calculate the product of the ion concentrations:
[Ag+] x [Cl-] = (0.00855 mol) x (10^-3 mol) = 8.55 x 10^-6
The solubility product constant (Ksp) of AgCl is typically around 1.8 x 10^-10 at room temperature. Since we assumed all activity coefficients to be equal to one, we can directly compare the product of ion concentrations to Ksp.
Since (8.55 x 10^-6) is greater than (1.8 x 10^-10), it means that the product exceeds the Ksp, indicating that AgCl will precipitate from the solution.
To calculate the weight of AgCl that forms per liter, we need to convert the moles of AgCl formed to grams using the molar mass of AgCl. The molar mass of AgCl is 143.32 g/mol, so:
Weight of AgCl = (0.00855 mol) x (143.32 g/mol) = 1.225 g
Therefore, 1.225 grams of AgCl will form per liter of solution.