A stock solution is prepared by adding 20mL of 0.2M MgF2 to enough water to make 80mL. What is the F– concentration of 20 mL of the stock solution?

The new soln has a concn of 0.2M x (20/80) = ?M MgF2. The fluoride is 2x that.

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To find the F- concentration of 20 mL of the stock solution, we can use the concept of stoichiometry and the dilution formula.

First, let's determine the number of moles of MgF2 in the 20 mL of the stock solution. We can use the formula:

moles = concentration * volume

In this case, the concentration of MgF2 is 0.2 M and the volume is 20 mL. We need to convert the volume from mL to liters by dividing by 1000:

volume = 20 mL / 1000 = 0.02 L

Now we can calculate the moles:

moles = 0.2 M * 0.02 L = 0.004 mol

Since MgF2 is a 1:2 compound, 1 mole of MgF2 will produce 2 moles of F-. Therefore, the number of moles of F- in the stock solution is:

moles of F- = 2 * 0.004 mol = 0.008 mol

Next, we need to determine the volume of the stock solution that contains these 0.008 moles of F-. The total volume of the stock solution is 80 mL, so the volume that contains 0.008 moles can be calculated using the dilution formula:

volume (F-) = (mol (F-) / mol (stock)) * volume (stock)

In this case, mol (F-) is 0.008 mol, mol (stock) is 0.004 mol, and volume (stock) is 20 mL. Again, we need to convert the volumes to liters:

volume (F-) = (0.008 mol / 0.004 mol) * (20 mL / 1000) = 0.016 L

Finally, we can calculate the concentration of F-:

concentration of F- = moles / volume

concentration of F- = 0.008 mol / 0.016 L = 0.5 M

Therefore, the F- concentration of 20 mL of the stock solution is 0.5 M.