Sally is goint to determine the quantity of MgSO4 by titrating the Mg2+ ions using EDTA. What is the concentration of Mg2+ (in mmol/L) in the sample if it takes 22.9 mL of EDTA solution with concentration of 0.01108 mol/L to titrate 50.00 mL of the sample to the endpoint? Note: each molecule of EDTA will complex with one Mg2+ion.

mL x M = millimoles Mg^+2.

Then M Mg^+2 = millimoles/50

To find the concentration of Mg2+ in the sample, we need to use the concept of stoichiometry and the volume and concentration information provided.

First, let's calculate the number of moles of EDTA that react with the Mg2+ ions in the sample. We know that it takes 22.9 mL of EDTA solution with a concentration of 0.01108 mol/L to titrate 50.00 mL of the sample.

Using the formula: moles = concentration × volume
moles of EDTA = 0.01108 mol/L × 0.0229 L
= 0.000253252 mol

Since each molecule of EDTA reacts with one Mg2+ ion, the number of moles of Mg2+ ions in the sample would be the same as the number of moles of EDTA used.

Next, we need to determine the number of moles of Mg2+ ions in the 50.00 mL of the sample. To do this, we will use the formula: moles = concentration × volume

moles of Mg2+ ions = concentration of Mg2+ × volume of the sample
= concentration of Mg2+ × 0.050 L

Now we can substitute the calculated value of moles of Mg2+ ions and the known volume of the sample to find the concentration of Mg2+ ions.

0.000253252 mol = concentration of Mg2+ × 0.050 L

Solving for the concentration of Mg2+ gives:

concentration of Mg2+ = 0.000253252 mol / 0.050 L
= 0.00506504 mol/L

Finally, we need to convert the concentration from mol/L to mmol/L by multiplying by 1000:

concentration of Mg2+ = 0.00506504 mol/L × 1000
= 5.06504 mmol/L

Therefore, the concentration of Mg2+ in the sample is 5.06504 mmol/L.