If the standard calcium carbonate solution is 100 ppm and 25 mL of solution is used for the standardization, calculate the approximate volume of 2mM EDTA solution needed to reach the end point of the titration.

I know to use M1V1=M2V2 but need help converting units of concentration from ppm to molar.

If it is 100 ppm that is 100 part is 1E6 parts or 1 part in 1E4 parts (see what I did?) or 0.1 in 1000 or 0.01 in 100 so how many would that be in 25?

0.0025?

right

To convert the concentration from ppm (parts per million) to molar, you need to know the molar mass of calcium carbonate.

Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 x 16.00 g/mol) (O) = 100.09 g/mol

First, convert the ppm concentration to grams per liter (g/L).

100 ppm = 100 mg/L (ppm is equivalent to mg/L)

1 ppm = 1 mg/L

Therefore, the concentration of the standard calcium carbonate solution is 100 mg/L or 100 mg/L * (1 g/1000 mg) = 0.1 g/L

Next, convert the concentration from g/L to molar:

0.1 g/L / 100.09 g/mol = 0.000999 mol/L = 0.000999 M

Using the formula M1V1 = M2V2, you can calculate the volume of the 2mM EDTA solution needed to reach the end point:

M1 = 0.000999 M
V1 = 25 mL = 0.025 L (you need to convert mL to L)
M2 = 2 mM = 0.002 M (you need to convert mM to M)
V2 = ?

0.000999 M * 0.025 L = 0.002 M * V2

V2 = (0.000999 M * 0.025 L) / 0.002 M

V2 = 0.0124875 L = 12.49 mL

Therefore, approximately 12.49 mL of the 2mM EDTA solution is needed to reach the end point of the titration.