A 30.00mL sample of a clear saturated solution of PbI2 requires 14.7mL of a certain AgNO3 for its titration.

I^-(saturated PbI2)+Ag^+(from AgNO3)--> AgI(s)

What is the molarity of this AgNO3?

I'm not really sure where to start.

Ksp PbI2 = about 1E-10 but you need to look it up.

Let x = solubility
...........PbI2 ==> Pb^2+ + 2I^-
............x........x.......2x
Ksp = (Pb^2+)(I^-)^2
1E-10 = (x)(2x)^2
Solve for x = solubility

Let's work backwards.

M = moles/L
Do we know L AgNO3? Yes, 14.7 mL or 0.0147 L. Do we know mole? No. Where can we get that? Look at the PbI2 (that's all that's left. right?)
PbI2(sat'd) ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2 = Ksp. You know Ksp, You can calculate solubility PbI2 and from that the (Pb^2+) as well as the (I^-). How many mols I^- do you have? That's mols x L (30 mL sample) = ?
The titration step is Ag^+ + I^- ==> AfI
mols I^- = moles Ag^+ from the balanced equation which gives you the moles you need back at the top. Solve for M Ag^+

how can i calculate the solubility with the information given ? in class we use concentrations to find solubility

So i found the solubility to be x=0.001296, is this the concentration of I^-? can i multiply this by 30mL to get my moles?

To find the molarity of AgNO3, you need to use the information provided in the question. The first step is to write a balanced equation for the reaction between PbI2 and AgNO3:

PbI2 + 2AgNO3 → 2AgI + Pb(NO3)2

From the balanced equation, you can see that it takes two moles of AgNO3 to react with one mole of PbI2.

Now, let's use the information given in the question. The 30.00 mL of saturated PbI2 reacted with 14.7 mL of AgNO3.

To determine the number of moles of AgNO3 used in the reaction, you'll need to know the molarity of AgNO3.

First, convert the volume of AgNO3 used (14.7 mL) to liters by dividing it by 1000:

14.7 mL ÷ 1000 = 0.0147 L

Next, you'll need to calculate the number of moles of AgNO3 used using its molarity. However, since we don't know the molarity of AgNO3 yet, let's call it "x".

moles of AgNO3 = molarity × volume in liters

moles of AgNO3 = x × 0.0147

Now, for the reaction between PbI2 and AgNO3, it takes two moles of AgNO3 to react with one mole of PbI2. Therefore, the number of moles of AgNO3 used is half the number of moles of PbI2 reacted.

moles of AgNO3 used = (1/2) × moles of PbI2

moles of AgNO3 used = (1/2) × moles of PbI2

The volume of PbI2 used is 30.00 mL, which is also equivalent to 0.03000 L. To find the moles of PbI2 used, you need to multiply its volume by its molarity.

moles of PbI2 = molarity × volume in liters

moles of PbI2 = ? × 0.03000

Since we don't know the molarity of PbI2, let's call it "y".

Now, you can set up an equation based on the stoichiometry of the reaction:

(1/2) × moles of PbI2 = moles of AgNO3

(1/2) × (molarity of PbI2 × 0.03000) = (molarity of AgNO3 × 0.0147)

Simplifying the equation:

(1/2) × (y × 0.03000) = (x × 0.0147)

0.015y = 0.0147x

Now, you have one equation with two unknowns. However, you can simplify further by substituting the molar mass of each compound.

The molar mass of PbI2 is 461.00 g/mol, and the molar mass of AgNO3 is 169.87 g/mol.

Substituting the molar masses into the equation:

(1/2) × (461.00 × 0.03000) = (169.87 × 0.0147)

Now, you have one equation with one unknown:

6.915 = 2.496x

Solving for "x" (the molarity of AgNO3):

x = 6.915/2.496

x ≈ 2.77 M

Therefore, the molarity of the AgNO3 solution is approximately 2.77 M.