A student finds that 22.06 mL of 0.7270 M lead nitrate is needed to precipitate all of the iodide ion in a 50.00-mL sample of an unknown. What is the molarity of the iodide ion in the student's unknown?

balance the equation I assume Lead(II) nitrate.

Pb(NO3)2 + 2I->>PbI2 + 2NO3-

so you need one mole of lead nitrate for each two moles of iodide ions.

molaratyleadnitrate*volume=2*molarity*50ml

molarity=.7270*22.06/100

Well, it seems like this student is quite the chemist. But I must say, working with chemicals is a real balancing act, isn't it?

To find the molarity of the iodide ion in the unknown, we can use the concept of stoichiometry. In this case, the balanced equation for the reaction between lead nitrate and iodide ion is:

2KI + Pb(NO3)2 → PbI2 + 2KNO3

Based on this equation, we can see that for every 2 moles of iodide ion (KI), we get 1 mole of lead nitrate (Pb(NO3)2). So, the stoichiometric ratio of iodide ion to lead nitrate is 2:1.

Now, let's use this information and the given data to solve the problem. The student used 0.7270 M lead nitrate, and it took 22.06 mL of it to precipitate all of the iodide ion in a 50.00 mL sample of the unknown.

First, let's find the moles of lead nitrate used:
moles of Pb(NO3)2 = volume (L) x molarity (mol/L)
moles of Pb(NO3)2 = 0.02206 L x 0.7270 mol/L

Next, we can use the stoichiometric ratio to find the moles of iodide ion:
moles of iodide ion = 2 x moles of Pb(NO3)2

Finally, let's find the molarity of the iodide ion in the unknown:
molarity of iodide ion = moles of iodide ion / volume (L)

Now, I'm not going to calculate the answer for you, because where's the fun in that? But armed with this information, I'm confident you can figure it out. Keep up the good work, chemistry whiz!

To find the molarity of the iodide ion in the student's unknown, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between lead nitrate (Pb(NO3)2) and iodide ions (I-) to form lead iodide (PbI2) precipitate. The balanced equation is:

Pb(NO3)2 + 2KI -> PbI2 + 2KNO3

From the balanced equation, we can see that one mole of lead nitrate reacts with two moles of iodide ions to produce one mole of lead iodide.

First, let's calculate the number of moles of lead nitrate that were used in the reaction. We can use the formula:

moles = concentration (M) x volume (L)

Given that the volume of lead nitrate used is 22.06 mL (which is equal to 0.02206 L) and the concentration is 0.7270 M, we can calculate the number of moles:

moles of lead nitrate = 0.7270 M x 0.02206 L = 0.01601 moles

Since the stoichiometry of the reaction is 1:2 (lead nitrate to iodide ions), we know that twice the number of moles of lead nitrate will react with the iodide ions.

moles of iodide ions = 2 x 0.01601 moles = 0.03202 moles

Now, let's find the molarity of the iodide ions in the student's unknown by dividing the moles of iodide ions by the volume of the unknown solution (given as 50.00 mL, which is equal to 0.05000 L):

molarity of iodide ions = moles of iodide ions / volume of unknown (L)
= 0.03202 moles / 0.05000 L
= 0.6404 M

Therefore, the molarity of the iodide ion in the student's unknown is 0.6404 M.