Suppose that a 10-mL sample of a solution is to be tested for Cl- ion by addition of 1 drop (0.2mL ) of 0.15 M AgNO3 .

What is the minimum number of grams of Cl- that must be present in order for AgCl(s) to form?

AgCl--->Ag^(+)+Cl^(-)

AgNO3= 2E^-4L*0.15mol/L
AgNO3/(.01L)=mol/L AgNO3=mol/L Ag^(+)=.003M

Find Ksp(AgCl) in solubility constant table
Ksp(AgCl)=1.77E-10
1.77E-10=[.oo3M Ag(+)][Cl^(-)]
solve for Cl^(-)

Cl^(-)=5.9E-8M
5.9E-8mol/L*.01L in tank * mm 35.453 g/mol=2*10^-8 g

AgCl ==> Ag^+ + Cl^-

Ksp = (Ag^+)(Cl^-)
You know 0.2 mL x 0.15 M AgNO3 will be moles of M x L. That divided by 10 mL (convert to L) will be the concn of Ag^+. Solve for (Cl^-) which will be in moles/L, convert that to moles/10 mL and convert moles Cl^- to grams Cl^-.

1e-7

Well, I usually clown around, but let me get serious for a moment. To figure out the minimum number of grams of Cl- required for AgCl(s) to form, we need to use stoichiometry.

First, let's convert the volume of the drop into liters: 0.2 mL is equal to 0.0002 L.

Now, we can use the concentration and volume of AgNO3 to find the number of moles of Ag+ ions in 0.2 mL of the solution:

moles of Ag+ = concentration * volume = 0.15 M * 0.0002 L

Next, we know that 1 Ag+ ion reacts with 1 Cl- ion to form AgCl(s). This means that the number of moles of Cl- ions must be the same as the number of moles of Ag+ ions.

So, the minimum number of grams of Cl- required for AgCl(s) to form would be equal to the moles of Cl- ions, which is the moles of Ag+ ions calculated above.

To convert moles into grams, we need to know the molar mass of Cl-. Since Cl- is part of NaCl (table salt), the molar mass of Cl- is approximately 35.45 g/mol.

Therefore, the minimum number of grams of Cl- required for AgCl(s) to form would be:

grams of Cl- = moles of Cl- ions * molar mass of Cl-
= moles of Ag+ ions * molar mass of Cl-

Hope that helps!

To find the minimum number of grams of Cl- that must be present in order for AgCl(s) to form, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between AgNO3 and Cl-. Here's how you can do it:

Step 1: Write the balanced chemical equation for the reaction between AgNO3 and Cl-.
AgNO3 + Cl- → AgCl(s) + NO3-

Step 2: Calculate the moles of AgNO3 used.
Since we are adding 1 drop (0.2mL) of 0.15 M AgNO3, we can calculate the moles of AgNO3 as follows:
0.2 mL × (1 L / 1000 mL) × (0.15 mol / L) = 0.00003 mol

Step 3: Determine the mole ratio between AgNO3 and Cl- from the balanced equation.
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of Cl- to form 1 mole of AgCl.

Step 4: Calculate the moles of Cl- required.
Since the mole ratio between AgNO3 and Cl- is 1:1, the moles of Cl- required will be the same as the moles of AgNO3 used: 0.00003 mol.

Step 5: Calculate the molar mass of Cl-.
The molar mass of Cl- is approximately 35.45 g/mol (from the periodic table).

Step 6: Calculate the minimum number of grams of Cl-.
To calculate the minimum number of grams of Cl-, multiply the moles of Cl- by its molar mass:
0.00003 mol × 35.45 g/mol = 0.001 grams

Therefore, the minimum number of grams of Cl- that must be present in order for AgCl(s) to form is 0.001 grams.