How many milliliters of a solution of 0.0013 M EDTA4- would be used to titrate the lead in 1000. mL of a 0.0073 M solution of Pb(NO3)2?

I'm sry, but I keep getting 5200ml and I know that's just wrong! >< can someone please help me?

OPPS ha, I got it, nvm

1000 x (0.0073/0.0013) = ?

You are allowed an answer with 2 significant figures.
I have an answer on my calculator of 5615.3 mL which rounds to 5600 or 5.6E2 to two s.f.
Check my work.

yes, that's it

To find the volume of EDTA4- solution needed to titrate the lead in the Pb(NO3)2 solution, you can use the concept of stoichiometry and the equation of the reaction between EDTA4- and Pb2+.

The balanced equation for the reaction is:
Pb2+ + EDTA4- -> Pb(EDTA)2-

From the equation, you can see that it takes one mole of Pb2+ to react with one mole of EDTA4- to form one mole of Pb(EDTA)2-.

First, you need to determine the number of moles of Pb2+ in the 1000 mL of the Pb(NO3)2 solution.
To do this, you can use the formula:

moles = concentration (M) x volume (L)

moles of Pb2+ = 0.0073 M x (1000 mL / 1000) L = 0.0073 mol

According to the stoichiometry of the reaction, it takes an equal number of moles of EDTA4- to react with Pb2+. Therefore, you need 0.0073 moles of EDTA4- to react with the Pb2+ ions.

Next, you need to find the volume of the 0.0013 M EDTA4- solution that contains 0.0073 moles of EDTA4-. You can use the following formula:

volume (L) = moles / concentration (M)

volume of EDTA4- solution = 0.0073 mol / 0.0013 M = 5.6 L

However, notice that the volume of EDTA4- solution is given in liters, not milliliters. To convert liters to milliliters, multiply the volume by 1000:

volume of EDTA4- solution = 5.6 L x 1000 mL/L = 5600 mL

Therefore, you would need 5600 mL (or 5.6 L) of the 0.0013 M EDTA4- solution to titrate the lead in 1000 mL of the 0.0073 M Pb(NO3)2 solution.