Calculate the mass of solid AgNo3 that can be added to 2.0 L of 0.10M K2CrO4 solution in order to just start precipitation?
Ag2CrO4 ==> 2Ag^+ + CrO4^2-
Ksp = (Ag^)^2(CrO4^2-)
mols CrO4^2- = mols x L = ?
Substitute into Ksp expression and solve for (Ag^+). That is (AgNO3). Convert M to mols in the 2.0 L and convert to grams. g = mols x molar mass. Post your work if you get stuck.
2AgNO3(aq) + K2CrO4(aq) ==> Ag2CrO4(s) + 2KNO3(aq)
Ag2CrO4(s) ===> 2Ag^+ CrO4^^2-
Ksp = [Ag^+]^2[CrO4^2-]
Ksp = (2x)^2(0.1)
To calculate the mass of solid AgNO3 that can be added to the K2CrO4 solution to just start precipitation, we need to use the concept of molar ratios and determine the limiting reagent.
Step 1: Write the balanced chemical equation for the precipitation reaction:
2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3
Step 2: Determine the molar ratios between the reactants and products from the balanced chemical equation:
From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of K2CrO4 to produce 1 mole of Ag2CrO4.
Step 3: Calculate the moles of K2CrO4 in 2.0 L of 0.10 M solution:
Molarity (M) = Moles of solute / Volume of solution (L)
0.10 M = Moles of K2CrO4 / 2.0 L
Moles of K2CrO4 = 0.10 M x 2.0 L = 0.20 mol
Step 4: Determine the moles of AgNO3 required using the mole ratio:
From the balanced equation, we know that 2 moles of AgNO3 react with 1 mole of K2CrO4. Therefore, we need twice the moles of K2CrO4 of AgNO3 to ensure complete reaction.
Moles of AgNO3 = 2 x Moles of K2CrO4 = 2 x 0.20 mol = 0.40 mol
Step 5: Convert moles of AgNO3 to mass using the molar mass:
The molar mass of AgNO3 = (1 x atomic mass of Ag) + (1 x atomic mass of N) + (3 x atomic mass of O)
= (1 x 107.87 g/mol) + (1 x 14.01 g/mol) + (3 x 16.00 g/mol)
= 169.87 g/mol
Mass of AgNO3 = Moles of AgNO3 x Molar mass of AgNO3
= 0.40 mol x 169.87 g/mol
= 67.948 g (rounded to three decimal places)
Therefore, the mass of solid AgNO3 that can be added to the 2.0 L of 0.10 M K2CrO4 solution in order to just start precipitation is approximately 67.948 grams.
To calculate the mass of solid AgNO3 needed to just start precipitation in the given solution, you need to use the concept of stoichiometry and the solubility rules of the compounds involved.
1. Write the balanced chemical equation for the reaction between AgNO3 and K2CrO4:
AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
2. Determine the stoichiometric ratio between AgNO3 and Ag2CrO4:
From the balanced equation, 1 mole of AgNO3 reacts with 1 mole of Ag2CrO4.
3. Calculate the concentration of Ag+ ions in the solution:
Since AgNO3 is soluble, it will dissociate fully in water, giving 1 Ag+ ion per formula unit of AgNO3.
The concentration of Ag+ ions is the same as the concentration of AgNO3, which is 0.10 M.
4. Determine the solubility of Ag2CrO4 using the solubility rules:
According to the solubility rules, Ag2CrO4 is insoluble and will form a precipitate.
5. Use the stoichiometry and concentration information to calculate the mass of AgNO3:
The molar mass of AgNO3 is 169.87 g/mol.
Mass of AgNO3 = (Concentration of Ag+ ions) x (Volume of solution) x (Molar mass of AgNO3)
= 0.10 M x 2.0 L x 169.87 g/mol
Calculate the result using the given values and obtain the mass of solid AgNO3 needed to just start precipitation.