5 drops of AgNO3 must be added to 1 L of a water sample: 46.1 mg SO42-/L and 30.6 mg Cl -/L.
What will be the precipitate that forms?
the molarity of AgNO3 is 0.0010 M (mol/L)
1 drop = 0.05 mL
Look up Ksp for Ag2SO4.
Look up Ksp for AgCl.
5 drops x (0.05 mL/drop) = mL added.
Convert to L, then M x L = moles added to the 1 L water and that will be the molarity.
Now calculate Qsp for Ag2SO4 and Qsp for AgCl and see if Qsp exceeds Ksp for either salt. If so, a ppt will form.
To determine the precipitate that forms when 5 drops of AgNO3 are added to the water sample, we need to consider the reactions between AgNO3 and the ions present in the water solution, specifically SO42- and Cl-.
First, let's convert the volume of AgNO3 from drops to liters using the given conversion factor.
1 drop = 0.05 mL
5 drops = 5 * 0.05 mL = 0.25 mL
Now, convert the volume of AgNO3 from milliliters to liters:
0.25 mL = 0.25 * (1 L / 1000 mL) = 0.00025 L
Next, we need to calculate the number of moles of AgNO3 added to the water solution. We can use the given molarity of AgNO3 to do this:
Molarity (M) = moles (mol) / Volume (L)
Rearranging the equation, we have:
moles (mol) = Molarity (M) * Volume (L)
moles of AgNO3 = 0.0010 M * 0.00025 L = 0.00000025 mol
Now, we can calculate the number of moles of SO42- and Cl- ions in the water sample, using their given concentrations:
For SO42-:
mass (mg) = concentration (mg/L) * Volume (L)
mass (g) = mass (mg) / 1000
mass of SO42- = 46.1 mg/L * 1 L / 1000 = 0.0461 g/L
moles of SO42- = mass (g) / molar mass (g/mol) of SO42-
The molar mass of SO42- is 96.06 g/mol.
moles of SO42- = 0.0461 g / 96.06 g/mol = 0.0004797 mol
For Cl-:
mass of Cl- = 30.6 mg/L * 1 L / 1000 = 0.0306 g/L
moles of Cl- = mass (g) / molar mass (g/mol) of Cl-
The molar mass of Cl- is 35.45 g/mol.
moles of Cl- = 0.0306 g / 35.45 g/mol = 0.0008642 mol
Now, let's determine the ratio of moles between AgNO3 and the ions present in the water sample:
AgNO3 : SO42- = 0.00000025 mol : 0.0004797 mol
AgNO3 : Cl- = 0.00000025 mol : 0.0008642 mol
From the ratio, we can see that the ratio between AgNO3 and Cl- is larger than the ratio between AgNO3 and SO42-.
This indicates that Cl- ions will react with Ag+ ions from AgNO3 to form a precipitate.
Therefore, the precipitate that forms when 5 drops of AgNO3 are added to the water sample is AgCl (silver chloride).