A 25.00 mL sample of a clear saturated solution of PbI2 requires 14.3 mL of a certain AgNO3(aq) for its titration.
I^-(from satd PbI2)+ Ag^+(from AgNO3)= AgI(s)
So, this is where I am.
using the Ksp equation for PbI2:
7.1*10^-9=4x^3
solving for x gives me 0.00121
Giving the concentration for I = 0.00242mol/L
Times this by 25ml gives me moles...
This is where I'm stuck
Sorry the question is What is the molarity of this AgNO3(aq)?
Never mind I got it, just divide by the other ml amount :)
To calculate the number of moles of I⁻ present in the 25.00 mL sample of the saturated solution, you need to multiply the concentration of I⁻ (0.00242 mol/L) by the volume of the solution (in liters).
First, convert the volume of the solution from milliliters to liters by dividing it by 1000:
25.00 mL ÷ 1000 mL/L = 0.02500 L
Next, multiply the concentration of I⁻ by the volume of the solution:
0.00242 mol/L × 0.02500 L = 0.0000605 mol
Therefore, there are 0.0000605 moles of I⁻ in the 25.00 mL sample of the saturated solution.
To continue from where you left off, you are on the right track. You have the concentration of I^- ions as 0.00242 mol/L. Now, you need to calculate the number of moles of I^- ions in the 25.00 mL sample. Here's how you can do that:
1. Convert the volume from milliliters to liters:
25.00 mL = 25.00 mL/1000 mL/L = 0.02500 L
2. Calculate the moles of I^- ions in the sample:
Moles of I^- ions = concentration of I^- ions x volume in liters
Moles of I^- ions = 0.00242 mol/L x 0.02500 L
Calculate this multiplication to find the number of moles of I^- ions in the 25.00 mL sample.
Once you have obtained the number of moles of I^- ions in the sample, you can use the stoichiometry of the reaction to find the number of moles of Ag^+ ions reacted.
From the balanced equation:
1 mole of PbI2 produces 2 moles of I^- ions
1 mole of AgNO3 produces 1 mole of Ag^+ ions
Since the molar ratio between I^- ions and Ag^+ ions is 1:1, the number of moles of Ag^+ ions reacted will be equal to the number of moles of I^- ions in the 25.00 mL sample.
I hope this helps! Let me know if you have any further questions.